T.C.

MANİSA CELAL BAYAR UNIVERSITY

PHYSICS I

LABORATORY MANUALS

2016

EXPERIMENT 1: ONE-DIMENSIONAL MOTION

OBJECTIVE: Study of one-dimensional rectilinear and uniformly accelerated motion.

1.1. UNIFORMLY ACCELERATED MOTION

THEORY

It is called the motion with constant acceleration if the change in velocity during equal time

intervals is always same for an object which moves along a straight line. Average acceleration is

equal to instantaneous acceleration of the object when the acceleration is constant. In this type of

motion, the velocity increases or decreases at the same amount from beginning to end of motion. If

we take ti=0 and tf=t, then the following equation can be written

𝐚𝐱=𝐯𝐱𝐬−𝐯𝐱𝐢

𝐭

or

𝐯𝐱𝐬 = 𝐯𝐱𝐢 + 𝐚𝐱𝐭 ( constant ax) (1.1)

The displacement of the motion can be written as the function of time as follows

𝚫𝐱 = 𝐱𝐬 − 𝐱𝐢

𝐱𝐬 − 𝐱𝐢 = 𝐯𝐢𝐱 𝐭 +𝟏

𝟐𝐚𝐱𝐭

𝟐 (constant ax) (1.2)

Finally, we replace the value of t in Equation 1.1 into Equation 1.2, time independent velocity

equation which depends on displacement is obtained as follows

𝐯𝟐 − 𝐯𝟎𝟐 = 𝟐𝐚(𝐱 − 𝐱𝟎) (1.3)

The position-time, velocity-time and acceleration-time graph of the object which moves with

constant acceleration have been given in Figure 1.1.

Figure 1.1: Variation graph for position, velocity and acceleration with respect to time for the motion with constant

acceleration

Another example to the motion with constant acceleration is the motion of an object on an inclined

plane.

Figure 1.2: The external forces exerted on a block lying on a frictionless incline plane

𝐅𝐱⃗⃗ ⃗ = 𝐦𝐠𝐬𝐢𝐧𝛉�̂� (1.4)

�⃗⃗� = 𝐦𝐠𝐜𝐨𝐬𝛉�̂� (1.5)

�⃗⃗� = −𝐍′⃗⃗⃗⃗ (1.6)

The forces acting on an object which moves on an inclined plane without friction are as shown in

Figure 1.2. When an object which is as rest on an inclined plane is falling down, it speeds up in the

direction of force and with the proportional to the magnitude of the acting force according to

Newton’s Second Law that is it makes a motion with acceleration.

A general relation between the force acting on an object and acceleration is given by

𝐅 = ∑𝐦�⃗� (1.7)

The acceleration of an object which moves on an inclined plane along x-direction is calculated by

𝐚𝐱⃗⃗⃗⃗ = 𝐠𝐬𝐢𝐧𝛉 �̂� (1.8)

Given the integral of acceleration with respect to time, velocity and position

𝐯𝐱⃗⃗ ⃗ = 𝐠𝐭 𝐬𝐢𝐧𝛉 �̂� (1.9)

�⃗� =𝟏

𝟐𝐠𝐭𝟐𝐬𝐢𝐧𝛉 �̂� (1.10)

On the above equations, it is assumed that the initial velocity of an object is zero.

THE EXPERIMENTAL PROCEDURE:

Incline the air table as shown in Figure 1.3 with the help of a wooden block.

Figure 1.3: Schematic drawing of an inclined air table

𝐬𝐢𝐧𝛂 =𝐡𝟐 − 𝐡𝟏

𝐋

Measure h1 and h2 which are the heights of the inclined plane’s both ends from the

bottom and L is the length of the inclined plane. Write the results to the table below.

Measured values

Period (ms)

𝐡𝟏(cm)

𝐡𝟐 (cm)

𝐋 (cm)

𝛂(0)

Place the carbon and measurement paper on to the surface of the air table.

Place one of the pucks on top edge of the air table in order to use it for measuring

and place the other puck to the top corner so that it doesn’t disturb the other’s motion

and then fix it at that place by a plastic fixer.

Do not touch the metal parts of the table during the measurements with spark.

Never switch on the spark if one or two pucks are out of table.

Do not keep one of the pucks at your hand when spark is on.

Otherwise you may get an harmless but disturbing electric shock.

o

Before starting the compressor, make fixed the air puck on top edge of the air table.

Run the air pump.

Switch on the spark timer and choose a frequency value either 40 or 60 or 80 Hz.

Press the red spark button in synchronization with the motion. Keep the button

pressed until the end of the motion

Switch off the air pump and spark timer to see your measurement results.

Pull out the paper on the surface of the air table and observe the burn points.

At least one of your measurements should be confirmed by the laboratory supervisor.

On the carbon paper there should be lots of burn points on the same line. While

choosing the starting point of your measurement you don’t need to take the first burn

point on the paper as the starting point of your measurement. Since the puck has a

small acceleration just before it set in motion?

Select a few more points from the reference point. Measure the distance between

points and then record the frequency value.

o The elapsed time between the two consequent points is calculated by the formula

t = 𝟏

𝐒𝐩𝐚𝐫𝐤 𝐅𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝐲

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CALCULATIONS

Record the ∆x and ∆𝐭 values to the table below.

∆xi(cm) ∆ti(s)

1

2

3

4

5

Use “gsinα” instead of “g” in all projectile motion equations since the projectile motion is

carried out with an inclined air table.

Calculate the theoretical acceleration value belongs to the motion with the formula;

𝐚𝐓 = 𝐠𝐬𝐢𝐧𝛉

Plot x–t2 graph according to the table above. If xi= 0 and vix = 0, the equation (1.2)

becomes

𝐱𝐬 =𝟏

𝟐𝐚𝐱𝐭

𝟐

where, 𝐚𝐱 = 𝟐𝐱 𝐭𝟐⁄

this value gives the slope of x–t2 graph and it is experimental acceleration.

Calculate the error for the result you obtained by using;

% 𝐄𝐫𝐫𝐨𝐫 =(𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 − 𝐄𝐱𝐩𝐞𝐫𝐢𝐦𝐞𝐧𝐭𝐚𝐥)

𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 × 𝟏𝟎𝟎

1.2. RECTILINEAR MOTION

THEORY

The motion of an object which moves with a constant velocity along a line is called rectilinear

motion. If the velocity of an object is constant, then its acceleration is zero. When the acceleration

on above equations is taken to be as a=0, the formula for uniform, linear motion is obtained

𝐱 = 𝐱𝟎 + 𝐯𝟎 𝐭 (1.12)

If we want to know the distance traveled by an object, we can write this formula in terms of

distance s as follows

𝐬 = 𝐱 − 𝐱𝟎 = 𝐯𝐭 (1.13)

Figure 1.4: Variation of the position, velocity and acceleration of a disc in a rectilinear motion

THE EXPERIMENTAL PROCEDURE

Level the air table off.

Place the carbon and measurement paper on to the surface of the air table.

Place one of the pucks on top edge of the air table in order to use it for measuring and place

the other puck to the top corner so that it doesn’t disturb the other’s motion and then fix it at

that place by a plastic fixer.

Run the air pump.

Switch on the spark timer and choose a frequency value either 40 or 60 or 80 ms.

Press the red spark button in synchronization with the projectile motion that you have

started with the launcher. Keep the button pressed until the end of the motion.

Switch off the air pump and spark timer to see your measurement results.

Pull out the paper on the bottom of the air puck and observe the burn points.

At least one of your measurements should be confirmed by the laboratory supervisior.

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On the carbon paper there should be lots of burn points on the same line. While choosing

the starting point of your measurement you don’t need to take the first burn point on the

paper as the starting point of your measurement. Since the puck has a small acceleration

just before set in motion.

Select a few more points from the reference point. Measure the distance between points and

then record the frequency value.

The elapsed time between the two consequent points is calculated by the formula

t = 𝟏

𝐒𝐩𝐚𝐫𝐤 𝐅𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝐲

CALCULATIONS

Record the ∆x and ∆t values to the table below.

∆xi(cm) ∆ti(s)

1

2

3

4

5

Plot x–t graph according to the table above. If xi= 0 , the equation (1.12) becomes

𝐱 = 𝐯𝟎 𝐭

where,

𝐯𝟎 = 𝐱 𝐭⁄

this value gives the slope of x–t graph and it is experimental velocity.

Calculate velocity value belong to the motion using following equation

𝐱 = 𝐯𝟎 𝐭

Calculate the error for the result you obtained by using;

% 𝐄𝐫𝐫𝐨𝐫 =(𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 − 𝐄𝐱𝐩𝐞𝐫𝐢𝐦𝐞𝐧𝐭𝐚𝐥)

𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 × 𝟏𝟎𝟎

QUESTIONS

1. Is it possible to have a case in which the acceleration of an object is zero if its velocity is not

zero? Explain.

2. Is the acceleration of an object different from zero if its velocity is zero? Explain.

3. An automobile moving with a speed of 75km/h is getting slow down by braking and its speed

getting fall down to 36 km/h in 5 seconds.

a) What is the constant acceleration of an automobile?

b) How much distance does an automobile travel during this time?

c) What is time passed to stop an automobile exactly from the beginning of the motion?

EXPERIMENT 2: PROJECTILE MOTION

OBJECTIVE: Study of a projectile motion on an inclined plane.

THEORY

PROJECTILE MOTION

Projectile motion of an object is simple to analyze if we make two assumptions: (1) the free-fall

acceleration is constant over the range of motion and is directed downward, and (2) the effect of air

resistance is negligible. With these assumptions, we find that the path of a projectile, which we call

its trajectory, is always a parabola as shown in Active Figure 2.1. We use these assumptions

throughout this experiment.

Figure 2.1: The parabolic path of a projectile that leaves the origin with a velocity v⃗ i. The velocity vector v⃗ changes

with time in both magnitude and direction. This change is the result of acceleration a⃗ = g⃗ in the negative y direction.

The expression for the position vector of the projectile as a function of time, with its acceleration

being that due to gravity, �⃗� = �⃗� .

𝐫 𝐬 = 𝐫 𝐢 + �⃗� 𝐢𝐭 +𝟏

𝟐�⃗� 𝐭𝟐 (2.1)

where the initial x and y components of the velocity of the projectile are

𝐯𝐱𝐢 = 𝐯𝐢𝐜𝐨𝐬𝛉𝐢 𝐯𝐲𝐢 = 𝐯𝐢𝐬𝐢𝐧𝛉𝐢 (2.2)

For a projectile launched from the origin, so that r i = 0. The final position of a particle can be

considered to be the superposition of its initial position r i ; the term v⃗ it, which is its displacement if

no acceleration were present; and the term 1

2g⃗ t2 that arises from its acceleration due to gravity. In

other words, if there were no gravitational acceleration, the particle would continue to move along

a straight path in the direction of v⃗ i . Therefore, the vertical distance 1

2g⃗ t2 through which the

particle “falls” off the straight-line path is the same distance that an object dropped from rest would

fall during the same time interval.

Two-dimensional motion with constant acceleration can be analyzed as a combination of two

independent motions in the x and y directions, with accelerations ax and ay. Projectile motion can

also be handled in this way, with zero acceleration in the x direction and a constant acceleration in

the y direction, ay = −g . Therefore, when analyzing projectile motion, model it to be the

superposition of two motions: (1) motion of a particle under constant velocity in the horizontal

direction and (2) motion of a particle under constant acceleration (free fall) in the vertical direction.

The horizontal and vertical components of a projectile’s motion are completely independent of

each other and can be handled separately, with time t as the common variable for both components.

HORIZONTAL RANGE AND MAXIMUM HEIGHT OF A PROJECTILE

Let us assume a projectile is launched from the origin at ti = 0 with a positive vyi component as

shown in Figure 2.2 and returns to the same horizontal level.

Figure 2.2: A projectile launched over a flat surface from the origin at ti = 0 with an initial velocity vi . The maximum

height of the projectile is h, and the horizontal range is R. At , the peak of the trajectory, the particle has coordinates

(R/2, h).

Two points in this motion are especially interesting to analyze: the peak point A, which has

Cartesian coordinates (R/2, h), and the point B, which has coordinates (R, 0). The distance R is

called the horizontal range of the projectile, and the distance h is its maximum height. Let us find

h and R mathematically in terms of vi, θi, and g.

We can determine h by noting that at the peak vyA = 0. Therefore:

𝐯𝐲𝐟 = 𝐯𝐲𝐢 + 𝐚𝐲𝐭

𝟎 = 𝐯𝐢𝐬𝐢𝐧𝛉𝐢 − 𝐠𝐭𝐀

𝐭𝐀 =𝐯𝐢𝐬𝐢𝐧𝛉𝐢

𝐠 (2.3)

Substituting this expression for tA into the y component of Equation 2.1 and replacing y = yA with

h, we obtain an expression for h in terms of the magnitude and direction of the initial velocity

vector:

𝐡 = (𝐯𝐢𝐬𝐢𝐧𝛉𝐢)𝐯𝐢𝐬𝐢𝐧𝛉𝐢

𝐠−

𝟏

𝟐𝐠 (

𝐯𝐢𝐬𝐢𝐧𝛉𝐢𝐠

)𝟐

𝐡 =𝐯𝐢𝟐𝐬𝐢𝐧𝟐𝛉𝐢

𝟐𝐠 (2.4)

The range R is the horizontal position of the projectile at a time that is twice the time at which it

reaches its peak, that is, at time tB = 2tA. Using the x component of Equation 2.1, noting that

vxi = vxB = vicosθi, and setting xB = R at t = 2tA, we find that

𝐑 = 𝐯𝐱𝐢𝐭𝐁 = (𝐯𝐢𝐜𝐨𝐬𝛉𝐢)𝟐𝐭𝐀

𝐑 = (𝐯𝐢𝐜𝐨𝐬𝛉𝐢)𝟐𝐯𝐢𝐬𝐢𝐧𝛉𝐢

𝐠=

𝟐𝐯𝐢𝟐𝐬𝐢𝐧𝛉𝐢𝐜𝐨𝐬𝛉𝐢

𝐠

Using the identity sin2θ = 2sinθcosθ, we can write R in the more compact form

𝐑 =𝐯𝐢𝟐𝐬𝐢𝐧𝟐𝛉𝐢

𝐠 (2.5)

INCLINED PLANE

The external forces exerted on a block lying on a frictionless incline plane is shown in Figure 2.3.

Figure 2.3: The external forces exerted on a block lying on a frictionless incline plane.

The net force acting on the block in y direction, according to the Newton’s law is:

𝐅 𝐲 = ∑𝐦�⃗� 𝐲 = 𝐦𝐠𝐬𝐢𝐧𝛂 �̂�

Hence the net acceleration of the block in y direction is:

𝐚𝐲 = 𝐠𝐬𝐢𝐧𝛂 (2.6)

When the projectile motion is studies on an inclined plane, the gravitational acceleration g in the

Equations 2.1 to 2.5 must be replaced by gsinα. That is:

𝐭𝐀 =𝐯𝐢𝐬𝐢𝐧𝛉𝐢

𝐠𝐬𝐢𝐧𝛂 (2.7)

𝐡 =𝐯𝐢𝟐𝐬𝐢𝐧𝟐𝛉𝐢

𝟐𝐠𝐬𝐢𝐧𝛂 (2.8)

𝐑 =𝐯𝐢𝟐𝐬𝐢𝐧𝟐𝛉𝐢

𝐠𝐬𝐢𝐧𝛂 (2.9)

THE EXPERIMENTAL PROCEDURE

Do not touch the metal parts of the table during the measurements with spark.

Never switch on the spark if one or two pucks are out of table.

Do not keep one of the pucks at your hand when spark is on.

Otherwise you may get an harmless but disturbing electric shock.

Incline the air table as shown in Figure 2.4 with the help of a wooden block.

Figure 2.4: Schematic drawing of an inclined air table

Measure the lengths h1, h2 and L of the inclined table. Write the results to the table below.

Place the carbon and measurement paper to the air table.

Place the air puck launcher to the BOTTOM RIGHT corner of the air table with an angle

you desire. Place the rubber band to the second position on the launcher.

Place one of the air puck to the launcher and fix the other puck to the top corner with the

help of a plastic fixer in such a way that it doesn’t block the motion of the other. (Figure

2.5).

Figure 2.5: Projectile method.

Run the air pump and try projectile motion few times until you get enough experience.

Switch on the spark timer and choose a frequency value (40,60 or 80 ms).

Press the red spark button in synchronization with the projectile motion that you have

started with the launcher. Keep the button pressed until the end of the motion.

Switch off the air pump and spark timer to see your measurement results.

The laboratory supervisor should confirm your chosen measurement when you are sure

about your measurement.

Draw the x and y coordinate projections as shown in Figure 2.6. Take the second

measurement point after starting point since the starting point can be problematic and place

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the corner of your coordinate axes to this point. Measure hmax and R values and record it to

the table below.

Figure 2.6: Measurement of the projectile motion data.

Determine t and tA flight times by counting the points on projectile motion. The time

elapsed between the two consequent points is ; T period (ms) .

Measure the lengths rsθ, ysθ, reθ, yeθ in order to determine the starting (θs) and the end (θe)

angles as shown in figure 2.7. Use more points as much as possible in such a way that the

linearity of the line is not violated. Extend the line in order to increase the precision of the

measurement rsθ, reθ.

Figure 2.7: Angle measurement of the initial and final velocities.

CALCULATIONS

Determine the angles θs and θe with the help of equations given below.

𝐬𝐢𝐧𝛉 =𝐲𝛉

𝐫𝛉 (2.10)

Determine the inclination angle of the air table by using the equation below.

𝐬𝐢𝐧𝛂 =𝐡𝟐−𝐡𝟏

𝐋 (2.11)

Use “gsinα” instead of “g” in all projectile motion equations since the projectile motion is

carried out with an inclined air table.

Figure 2.8: A sample projectile motion data view.

Take the average of the equal distances on x axes between the points.

Calculate the x component of the initial velocity vxi by using the equation below:

𝐯𝐱𝐢 =�̅�

𝐭 (2.11)

where t is the multiplication of the number of points N on paper and T period (ms) .This

gives us the flight time experimentally.

Calculate the initial velocity experimentally by using the equation 𝐯𝐬𝐢 =𝐯𝐱𝐢

𝐬𝐢𝐧𝛉𝐬=

𝐯𝐱𝐢𝐫𝐬𝛉

𝐲𝐬𝛉.

Calculate also the end velocity vei using the same equation.

Can 𝛉𝐬 and 𝛉𝐞 angles be different from each other? Make comments on the corresponding

reasons and the results.

Calculate the experimental value of the time to reach maximum height by using number of

points N:

𝐭𝐀 = 𝐍𝐱𝐓(𝐩𝐞𝐫𝐢𝐨𝐝) (2.12)

Calculate the theoretical value of the time to reach maximum height by using the equation

(2.7) .

Calculate the time t = 2tA to reach the range R experimentally.

Measure the experimental value of the range Rexp by using a ruler. Calculate the same value

theoretically by using the Equation 2.9.

Measure the experimental value of the maximum height by using a ruler. Calculate the

same value theoretically by using the Equation 2.8.

Write your results to the corresponding table below and make the error calculation for each

result you obtained by using;

% 𝐄𝐫𝐫𝐨𝐫 =(𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥− 𝐄𝐱𝐩𝐞𝐫𝐢𝐦𝐞𝐧𝐭𝐚𝐥)

𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 × 𝟏𝟎𝟎 (2.14)

MEASUREMENTS TABLE

Measured Values

T Period (ms)

𝐡𝟏(cm)

𝐡𝟐 (cm)

𝐋 (cm)

𝐫𝐬𝛉(cm)

𝐲𝐬𝛉 (cm)

𝐫𝐞𝛉(cm)

𝐲𝐞𝛉 (cm)

Time to reach maximum height (𝐭𝐀) (s)

Flight time t(s)

Range (R) (cm)

Maximum height (hmax) (cm)

RESULTS TABLE

𝐄𝐱𝐩𝐞𝐫𝐢𝐦𝐞𝐧𝐭𝐚𝐥 𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 % Error

𝛉𝐬 ( 0

)

𝛉𝐞 ( 0

)

α ( 0

)

𝐯𝐬𝐢 (cm/s)

𝐯𝐞𝐢 (cm/s)

Time to reach maximum height (𝐭𝐀) (s)

Flight time t(s)

Range (R) (cm)

Maximum height (hmax) (cm)

QUESTIONS

1. Which parameter remains constant for an object making a projectile motion ?

a. Magnitude of its velocity,

b. Acceleration,

c. Horizontal projection of its velocity,

d. Vertical projection of its velocity.

2. A ball hit by a soccer player has an initial velocity of 30 m/s and it makes 37o angle with

horizontal axes.

a. Write the equation of motion of the ball.

b. What is the position and velocity components of the ball at t=1s?

c. How long it takes to reach the maximum height?

d. What is the maximum height that ball reaches?

e. How far the ball goes before hitting the ground?

3. A ball is thrown away from the roof of a 33 m high building, with an angle of 53o with

horizontal and with an initial speed of 5 m/s. At ground, a children starts running

horizontally at the same time with a constant acceleration of a.

a. Write the equation of motion of the ball by choosing a coordinate system.

b. What is the flight time of the ball?

c. How far away from the building the ball goes?

d. What should be the acceleration a of the children to be able to catch the ball before

hitting the ground?

EXPERIMENT 3: DYNAMIC

PURPOSE: In this lab, we will explore the relationship between position, velocity and

acceleration. In this experiment, friction will be neglected. Constant (uniform) acceleration

will be investigated.

1.1. NEWTON’S LAW

INTRODUCTION

Speed is just how fast something is moving. It is measured in how far something moves in

some time. Speed is given by the equation:

Speed=x/t (3.1)

The symbol ’∆’ indicates a change is some quantity such as position (’x’) in this case. The

unit of speed is always distance divided by time. We might measure speed in miles/hour,

meters/second etc. In this lab we will use meters/second (m/s).

Speed is how far something moves in some time. Obviously this depends on the time period

we use. Things rarely move at constant speed. We measure the speed of something over a

very short time, we can talk about the instantaneous speed. Instantaneous speed is like

taking a snap shot of the speed. The average speed is what we have in equation 2.1 for

some time period that may not be very small.

Just as the position changing gives us velocity, when velocity changes we have an

acceleration. Acceleration is:

Acceleration= a = change in velocity / time interval = v/t (3.2)

Acceleration has units of distance divided by time. For this lab, we will use meter/second2

(m/s2)

When we specify a speed and the direction, we are giving a velocity. Velocity is a

vector. Like all vectors it has magnitude (how much) and direction (which way). Acceleration

being the change in velocity is also a vector. In this lab, we will restrict the motion to one

dimension so the vector nature of velocity and acceleration are no so important. However,

you should remember that velocity and acceleration are vectors.

If the acceleration is constant, there are relatively simple relationships between position,

velocity and acceleration in one dimension. If an object at time zero (t = 0) is at the origin

(x = 0), the the velocity and position are given by:

v(t) = v0 + a.t (3.3)

𝐱(𝐭) = 𝐯𝟎. 𝐭 +𝟏

𝟐𝐚. 𝐭𝟐 (3.4.)

where a is the constant acceleration, t is the time and v0 is the initial velocity of the object

at time zero. For this experiment, we will have motion in one dimension with zero initial

velocity.

Next we should define force. Force is simply a push or a pull. It can be from gravity

i.e. gravity pulls on a mass, a magnetic ’pull’ between a north and south magnetic pole or

simply you pushing a book across a table.

Newton’s 2nd law relates mass, acceleration and force. In words:

The acceleration of n object is directly proportional to the net force acting on the object, is in the

direction of the net force and is inversely proportional to the mass of the object.

F = m.a (3.5)

The units of force in the metric system are the newton.

There are several things we should notice about this statement. The equation for New-

ton’s 2nd law tells us force is a vector like velocity or acceleration. When you push

something, you push in a direction e.g, north or down, as well as some amount e.g. 10 N.

We can use Newton’s 2nd law to figure out what a newton (N) is in terms of mass, time

and distance. Since force is mass times acceleration, the units of a newton can be broken

down to kg m/s2 . Finally, notice the statement of Newton’s 2nd law says ’net force’. This

means all the forces added together acting on an object. Since they are vectors, we have

to add all the forces vectors.

We will investigate Newton’s second law and constant acceleration with the air table in

the configuration shown in figure 1.2. The hanging mass, m, will pull the puck across the

table as it falls. The mass of the puck is mp. The string connecting the hanging mass and

the puck has a tension, T. We assume there is no friction between the table and the puck.

The force on the puck is simply the tension (pull of the hanging mass). Assuming no

friction, Newton’s 2nd law gives:

Force = T = mp.a (3.6.)

Figure 3.1. Equipments of experiment.

a

mp T

T

a mg

Figure 3.2. The air table arrangement.

The forces on the hanging mass are T (upward) and the force of gravity, mg (downward). ’g’

is the acceleration of gravity (9.81 m/s2 ). The acceleration of the hanging mass is downward.

Newton’s second law gives:

Force = T – m.g = m (-a) (3.7)

If we substitute equation 2.6 into equation 2.7 and solve for the acceleration we have:

𝐚 =𝐦.𝐠

𝐦𝐩+𝐦 (3.8)

This equation shows that the acceleration is constant and proportional to the acceleration

of gravity through the ratio

PROCEDURE

Special Caution:

To avoid electrical shock, do not touch the pucks or air table while the spark

timer is on. Use an insulator such as a rolled up sheet of paper to push or hold

the puck.

The air table apparatus is shown in figure 2. The white box at the back of the

table is the spark unit. Two pucks are shown connected to the rubber hoses

which supply the air to float the puck.

Newton’s Second Law;

In this section of the procedure, the puck will be pulled across the table by a hanging mass

connected to the puck by a string over a pulley.

•Attach the pulley to the side of the table if it is not already attached. Connect the

string to the collar and the other end to the hanging mass. Place the string over the

pulley wheel.

•Using the unmarked side the white paper, configure the free puck on several folds of

the corner of the paper so it will not move. Hold the puck connected to the hanging

mass at the far end of the table with the insulated tube. Press the sparker pad and

release the puck. The puck should have zero initial speed so the puck should not be

released until the sparker has started. Stop the spark timer when the mass reaches the

end of the table.

•Turn off (for safety) the sparker power supply, remove both pucks, lift up the paper

sheet and turn it over to examine the sparker dots marking the puck trajectory.

•Qualitatively describe how the dots are spaced on the paper.

•Measure the distance between the first sparker dots another dot approximately 0.20

meters from the first dot. Count the number of time intervals (0.05 seconds between

dots) to determine the total time between the first dot and the dot where you measure

the distance. Using equation 2.4, calculate the acceleration assuming v0 = 0.

•Using equation 2.8, calculate the acceleration of gravity. Calculate the percentage error

for your value with the standard value of 9.81 m/s2

QUESTIONS

•In the constant acceleration part of the experiment (Newton’s 2nd law), if the initial

velocity is not zero would the final result be serious effected?

•In the constant acceleration part of the experiment (Newton’s 2nd law) there may be

some friction between the puck and air table. Assuming the friction is constant, would

you still expect the acceleration to be constant? Would it change the value of g you

calculated?

EXPERIMENT 4: MOMENTUM AND COLLISION

PURPOSE OF THE EXPERIMENT: For elastic collisions in an isolated system, examine

the conservation of linear momentum and kinetic energy.

ELASTIC COLLISION

GENERAL INFORMATION

Linear momentum of an object "P", is the product of mass and velocity

�⃗⃗� = 𝐦�⃗� (4.1)

Here we will talk about the momentum briefly from the linear momentum. However, only the

net external force when applied, we know that the speed of the object changed and this means

that the momentum change. This fact can be seen from Newton's second law. According to

Newton's second law for a constant mass of a body;

𝐅 𝐞𝐱𝐭 = 𝐦�⃗� = 𝐦𝐝�⃗�

𝐝𝐭 (4.2)

When m is constant, this equation is written as follows:

𝐅 𝐞𝐱𝐭 =𝐝(𝐦�⃗� )

𝐝𝐭=

𝐝�⃗⃗�

𝐝𝐭 (4.3)

From the above equation, if an external force acts on an object, the object's momentum does

not protected. That is, momentum does not change with time. If F⃗ ext = 0 then

𝐝�⃗⃗�

𝐝𝐭= 𝟎 (4.4)

�⃗⃗� = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 (4.5)

Here the momentum not change with time, the object always have the same momentum.

N- particle system consist of m1, m2,…….mN masses can be generalized based on the above

results. When we are dealing with a system of objects (m1, m2,…….mN), the total momentum

P⃗⃗ tot of the system is the vector sum of the individual momentums:

�⃗⃗� 𝐭𝐨𝐭 = �⃗⃗� 𝟏 + �⃗⃗� 𝟐 + ⋯+ �⃗⃗� 𝐍 (4.6)

Where are,

𝐏𝟏⃗⃗ ⃗⃗ ⃗ = 𝐦𝟏𝐯𝟏⃗⃗⃗⃗ , 𝐏𝟐⃗⃗⃗⃗ = 𝐦𝟐𝐯𝟐⃗⃗⃗⃗ ,….. 𝐞𝐭𝐜. (4.7)

The sum in the equation (4.6) is a vector sum process. In this situation, if a generalized

equation (4.3);

𝐅 𝐞𝐱𝐭 =𝐝�⃗⃗� 𝐭𝐨𝐭

𝐝𝐭=

𝐝

𝐝𝐭(𝐏𝟏⃗⃗⃗⃗ + 𝐏𝟐⃗⃗⃗⃗ + ⋯+ 𝐏𝐍⃗⃗ ⃗⃗ ) (4.8)

where Fext, the system comprised of the particles refers to the net external force. This external

forces may be friction and gravity. Hence in the system formed by particles it does not have

any total external force and the total momentum of the system will be protected. So;

𝐝�⃗⃗� 𝐭𝐨𝐭

𝐝𝐭=

𝐝

𝐝𝐭(�⃗⃗� 𝐭𝐨𝐭 = 𝐏𝟏⃗⃗⃗⃗ + 𝐏𝟐⃗⃗⃗⃗ + ⋯ �⃗⃗� 𝐍) = 𝟎 (4.9)

�⃗⃗� 𝐭𝐨𝐭 = �⃗⃗� 𝟏 + �⃗⃗� 𝟐 + ⋯+ �⃗⃗� 𝐍 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 (4.10)

Above collection is a vector sum process.

If the system of interest is isolated, that is not acted upon by an external force, the total

momentum of the system remains constant.

In this study, in the air table a horizontal position will be investigated momentum

conversation with two-pucks system. In the horizontal position, any external force does not

occur on the pucks on the air table which minimizes friction. Therefore, the total momentum

of the pucks seems to be preserved. Pucks are provided collisions, the total momentum before

and after the collision are measured and compared. The spots obtained in the experiment are

given in the following figure on data paper.

Figure 4.1 : Data points of the two magnetic pucks that performed elastic collisions on the air table in the

horizontal positions.

Velocities of two pucks are respectively v⃗ A , v⃗ B and v⃗ A′ , v⃗ B

′ before and after the collision.

Momentum is conversed because the system is isolated and at any time;

�⃗⃗� 𝐭𝐨𝐭 = 𝐜𝐨𝐧𝐬𝐭𝐚𝐧𝐭 (4.11)

�⃗⃗� 𝐀+ �⃗⃗� 𝐁 = �⃗⃗� 𝐀′ + �⃗⃗� 𝐁

′ (4.12)

Where the momentums are P⃗⃗ A = mAv⃗ A, P⃗⃗ B = mBv⃗ B. Because the masses of the pucks are

same, the above equation is converted as follows.

�⃗� 𝐀 + �⃗� 𝐁 = �⃗� 𝐀′ + 𝐯𝐁

′ (4.13)

It is shown method that the vector sum in the equation 4.13 (above) are geometrically found

in the experimental procedure section. Since the system is isolated, the momentum is

conserved at an inelastic collision. In such a collision, two pucks move sticking together, as

an object having a mass 2m and the velocity v⃗ ′. The points in the data sheet should resemble

the Figure 4.1.

Another concept that will be encountered in this experiment is the center of mass (CM). The

CM of symmetrical objects such as cubes (figure 4.2a) and spheres (figure 4.2b ) is the same

with the theirs geometric center. The CM of the shape in the figure 4. 2c is predictably the

midpoint of the rod.

Figure 4.2: CM of some symmetric homogeneous objects.

CM for mass distribution in different shapes should be redefined. CM of position vector R⃗⃗ of

a system with N particles is defined as follows, (seen Figure 4.2)

�⃗⃗� =𝐦𝟏𝐫 𝟏+𝐦𝟐𝐫 𝟐+⋯..+𝐦𝐍𝐫 𝐍

𝐦𝟏+𝐦𝟐+⋯+𝐦𝐍 (4.14)

Where r 1, r 2, … , r N are position vectors and m1, m2, … .mN are masses.

Figure4. 3: CM of R for mass distribution.

If the particles change their position with time, the position of the CM changes and the

vector exchange rate of CM considered as the center of mass velocity.

�⃗⃗� 𝐂𝐌 =𝐝�⃗⃗�

𝐝𝐭 (4.15)

when we take the derivative of both sides of equation (4.14) for particles with constant mass.

�⃗⃗� =𝐦𝟏𝐫 𝟏+𝐦𝟐𝐫 𝟐+⋯..+𝐦𝐍𝐫 𝐍

𝐦𝟏+𝐦𝟐+⋯+𝐦𝐍 (4.16)

�⃗⃗� 𝐂𝐌 =𝐦𝟏�⃗� 𝟏+𝐦𝟐�⃗� 𝟐+⋯+𝐦𝐍�⃗� 𝐍

𝐦𝟏+𝐦𝟐+⋯+𝐦𝐍 (4.17)

are obtained. The points in the equation (4.16) means derivative so that these are only

speeds. When the above equation is applied to the two pucks systems;

�⃗⃗� =𝐦𝐫𝐀⃗⃗⃗⃗ +𝐦𝐫𝐁⃗⃗⃗⃗

𝐦+𝐦 (4.18)

�⃗⃗� =𝐫𝐀⃗⃗⃗⃗ +𝐫𝐁⃗⃗⃗⃗

𝟐 (4.19)

are obtained. Since the masses of the pucks are equal here, equation 4.20 is obtained by

removing the masses. Thus velocity of CM,

�⃗⃗� 𝐂𝐌 =𝐯𝐀+𝐯𝐁

𝟐 (4.20)

There are some important consequences of the above equation. In two pucks sytem, firstly,

while maintaining momentum total on the right side of the equation are constant (compare

with equation 4.13). This situation means that velocity of CM is constant under these

conditions. In other words, the CM moves at a constant velocity. (Constant velocity means

that the magnitude and direction of the speed does not changes ). Thus CM of the system

always moves at a linear constant speed for a system isolated that the total momentum is

conserved. This situation also shows that the velocity is equal to half of total velocities of both

masses. Therefore, velocity equations are as follows for our two-pucks system, before and

after the collision.

�⃗⃗� 𝐂𝐌 = �⃗⃗� 𝐂𝐌′ (4.21)

𝐕𝐂𝐌=

�⃗� 𝐀+ �⃗� 𝐁𝟐

= �⃗⃗� 𝐂𝐌′ =

�⃗� 𝐀′ +�⃗� 𝐁

′

𝟐

(4.22)

In this experiment, It will be investigated kinetic energy conservation of the pucks for the

collision secondly. Let us remember the definition of kinetic energy K of an object that have

mass m and linear velocity v.

𝐊 =𝟏

𝟐𝐦𝐯𝟐 (4.23)

Therefore total kinetic energy of the two-pucks system prior to the elastic collision;

𝐊 =𝟏

𝟐𝐦𝐯𝐀

𝟐 +𝟏

𝟐𝐦𝐯𝐁

𝟐 (4.24)

and kinetic energy after the collision;

�́� =𝟏

𝟐𝐦𝐯𝐀

𝟐́ +𝟏

𝟐

́𝐦𝐯𝐁

𝟐́ (4.25)

However, the two pucks sticks each other in an inelastic collisions. After this collision, these

two pucks moves as an object that have mass 2m and velocity v. Thus ıts kinetic energy;

𝐊′ =𝟏

𝟐(𝟐𝐦)𝐯𝟐′ = 𝐦𝐯𝟐′ (4.26)

Since the kinetic energy is a scalar quantity, the total in equation (4.25) and (4.26) is a scalar

collection process. On the other hand, the kinetic energy is preserved in almost elastic

collision that is K = 𝐊′

EXPERIMENTAL PROCEDURE

Please run the pump switch (P).

Gently launch the two pucks diagonally towards each other so they closely

approach each other and repel without touching. Repeat this process several times

until sufficient degree appropriate get collision.

Now, set the period from the the spark generator (example 60 ms)

Later throw the puck to the other side by an air table when running the P key and

so run the of the spark generator as soon as the pucks remain constant

Hold both switches open until two pucks movement have completed

CALCULATIONS

Remove the data sheet and then please carefully review the resulting points. Spots

should be like in Figure 4.1. Points for each pucks 0, 1, 2, …. and so on the

numbering.

Two or three of the range of each, measure the length divided by the time on the

road. Later each puck collide rate found before and after the collision. Pucks come

in the way of naming the A and B before the collision and the A´ and B ´ after the

collision.

Find the vector sum ve . Example; Find the vector addition of

and . For example to lengthen A and B ways for finding

. After draw vector of this velocities size which is relation with the length

where start the crosssection and direction. For instance velocity of 10 m/s

can be drawn for 1cm length vector. Then, find the sum of this velocities using

parallelogram adding. Make the same method for .

Define the points that made at same time after and before collision. Specify the

location of the center of mass combining those points.

Find the speed using the obtained recording for CM, before and after the collision.

Find the kinetic energy of two-pucks before and after the collision and compare

them.

QUESTIONS

1. How to change the momentum and kinetic energy, if the velocity of a particle

doubles?

2. Are their momentums also equal, if the kinetic energy of two objects are equal?

Explain why.

3. As a result of the full elastic collision between two particles, does the kinetic energy

of each particle change?

4. Is it possible for a body, the center of mass is being outside of its actual volume? If

your answer is "Yes", give an example?

EXPERIMENT 5: INELASTIC COLLISIONS

OBJECTIVE: To investigate that in inelastic collisions the energy and momentum equations

THEORY

INELASTIC COLLISIONS

Inelastic collisions; where the two colliding bodies adhere to each other upon contact and

move off as a unit after the collision. Linear momentum is conserved, while not conserved

mechanical energy in inelastic collision. After the collision, system moves from its axis of

rotation, linear velocity and the speeds of the center of mass of both two bodies (m1, m2) are

equal for each one. The velocities of first (m1), and second (m2) are represent v1⃗⃗ ⃗, v2⃗⃗ ⃗ ve u1⃗⃗⃗⃗ , u2⃗⃗⃗⃗

for before and after collision. If the V⃗⃗ is represents and the speeds of the center of mass;

𝐮𝟏⃗⃗ ⃗⃗ = 𝐮𝟐⃗⃗ ⃗⃗ = �⃗⃗� (5.1)

can be written as (5.1). Since momentum is conserved in inelastic collisions;

𝐦𝟏⃗⃗⃗⃗⃗⃗ 𝐯𝟏⃗⃗⃗⃗ + 𝐦𝟐⃗⃗⃗⃗⃗⃗ 𝐯𝟐⃗⃗⃗⃗ = 𝐦𝟏⃗⃗⃗⃗⃗⃗ 𝐮𝟏⃗⃗ ⃗⃗ + 𝐦𝟐⃗⃗⃗⃗⃗⃗ 𝐮𝟐⃗⃗ ⃗⃗ = (𝐦𝟏 + 𝐦𝟐)�⃗⃗� (5.2)

the equations (5.2) can be written. The speed of the center of mass(V)⃗⃗⃗⃗ ⃗⃗ ; can derivative as following.

�⃗⃗� =𝐦𝟏𝐯𝟏⃗⃗⃗⃗ + 𝐦𝟐𝐯𝟐⃗⃗⃗⃗

𝐦𝟏 + 𝐦𝟐

(5.3)

If the masses m2 and m1are equal to each other the speed of the center of mass(V)⃗⃗⃗⃗ ⃗⃗ can be obtained (5.4);

�⃗⃗� =𝐯𝟏⃗⃗⃗⃗ + 𝐯𝟐⃗⃗⃗⃗

𝟐

(5.4)

Inelastic collision, the total momentum of the system although there is a change in the

system’s kinetic energy is always a loss. So, the equations of the kinetic energies for before

and after collision are written, respectively.

(𝟏

𝟐𝐦𝟏𝐯𝟏

𝟐 +𝟏

𝟐𝐦𝟐𝐯𝟐

𝟐) > (𝟏

𝟐(𝐦𝟏 + 𝐦𝟐)𝐕

𝟐) (5.5)

If the m1 = m2 the equation (5.5) converted as following;

(𝐯𝟏𝟐 + 𝐯𝟐

𝟐) > 𝟐𝐕𝟐 (5.6)

The energy difference is transformed into heat or other forms of energy in inelastic collisions.

If K1 and K2 are defined the total kinetic energy before the collisions, reduced rate of the

system can be expressed the following equation (5.7);

𝐞 =𝐊𝟏 − 𝐊𝟐

𝐊𝟏

(5.7)

THE EXPERIMENTAL PROCEDURE:

Wrap the special adhesive tape around the equal mass of two disks as outward facing

adhesive surfaces

After turning the horizontal air table, put the pucks which is close to the corners of the

table for you.

Switch on the spark timer and choose a T period value either 40 or 60 or 80 ms.

Press the air pedal, collision somewhere in the middle of the table makes you.

You will see that pucks act together after the collision. After you set up the positions

and velocity of the pucks, pressing air pedal, you repeat this motion.

You must press the pedal after the motion of pucks. If you press the pedal before the

motion, to find location or position (relative position in time t) of pucks can be

difficult.

Switch off the air pump and spark timer to see your measurement results.

Place the carbon and measurement paper on to the surface of the air table.

Pull out the paper on the surface of the air table and observe the burn points.

At least one of your measurements should be confirmed by the laboratory supervisor.

Do not touch the metal parts of the table during the measurements with spark.

Never switch on the spark if one or two pucks are out of table.

Do not keep one of the pucks at your hand when spark is on.

Otherwise you may get an harmless but disturbing electric shock.

http://tureng.com/search/laboratory

According to results, draw the vector and measure the magnitude of this vector using

by ruler, v1⃗⃗ ⃗, v2⃗⃗ ⃗ and u1⃗⃗⃗⃗ , u2⃗⃗⃗⃗ before and after collision. If the velocities of after the

collision, u1⃗⃗⃗⃗ ve u2⃗⃗⃗⃗ , are not equal, this system revolves around itself.

CALCULATIONS

Show the vector R⃗⃗ = v1⃗⃗ ⃗ + v2⃗⃗ ⃗ and R⃗⃗ ′ = u1⃗⃗⃗⃗ + u2⃗⃗⃗⃗ , drawing by ruler. Are equal to the

resultant vector (R⃗⃗ ) before the collision and resultant vector (R⃗⃗ ′) after the collision.

How can you show the conservation of linear momentum?

Draw the speed of the center of mass(V)⃗⃗⃗⃗ ⃗⃗ .

V⃗⃗ are equal velocities u1⃗⃗⃗⃗ ve u2⃗⃗⃗⃗ after the collision. How is there a equation between V⃗⃗

and v1⃗⃗ ⃗, v2⃗⃗ ⃗ . Measure the magnitude of velocity vector V⃗⃗ .

You use your results of velocity the equation (5.6) to show deprotection of kinetic

energy.

Find the reduced rate of kinetic energy using by equation (5.7).

Figure 5.1: Inelastic collisions

QUESTIONS

1. Give examples for the Inelastic collisions, encountered in our everyday life.

2. B If the velocity of a particle doubles, how the momentum changes? What happens to the

kinetic energy?

EXPERIMENT 6: SIMPLE HARMONIC MOTION

OBJECTIVE: To investigate Hooke's Law and simple harmonic motion. The potential,

kinetic energy and total energy conservation law to be tested in the system, simple harmonic

motion.

THEORY

SIMPLE HARMONIC MOTION

A fixed point in doing both sides of the oscillating object is called vibrational motion. In this

experiment you will explore simple harmonic motion, special status of vibrational motion.

Think a mass (m) attached, a spring clips attached to one end. If the system is left at rest at the

equilibrium position then there is no net force acting on the mass. However, if the mass is

displaced from the equilibrium position, a restoring elastic force which obeys Hooke's law is

exerted by the spring. Mathematically, the restoring force F is given by

𝐅 = −𝐤�⃗� (6.1)

where, F is the restoring elastic force exerted by the spring (in SI units: N), k is the spring

constant (N·m−1

), and x is the displacement from the equilibrium position (in m).

Displacement is proportional to the force and direction is always toward the equilibrium

position that is the opposite direction with displacement. Once the mass (m) is displaced from

its equilibrium position (x), it experiences a net restoring force. As a result, it accelerates and

starts going back to the equilibrium position. So net restoring force and acceleration of mass

can be expressing the following equations (6.2), (6.3);

𝐅 = −𝐤�⃗� = 𝐦𝐚 (6.2)

�⃗� = −𝐤

𝐦�⃗�

(6.3)

The acceleration of mass is proportional with displacement from the equilibrium position and

opposite direction of displacement. So, For one-dimensional simple harmonic motion, the

equation of motion, which is a second-order linear ordinary differential equation with constant

coefficients, could be obtained by means of Newton's second law and Hooke's law.

𝐝𝟐�⃗�

𝐝𝐭𝟐= −

𝐤

𝐦�⃗�

(6.4)

where m is the inertial mass of the oscillating body, x is its displacement from the equilibrium

(or mean) position, and k is the spring constant and , k

m= w2 , ve w, the angular frequency.

Solving the differential equation above, a solution which is a sinusoidal function is obtained.

𝐱(𝐭) = 𝐀𝐜𝐨𝐬(𝐰𝐭 + 𝛗) (6.5)

In the solution, c1 and c2 are two constants determined by the initial conditions, and the

origin is set to be the equilibrium position. Each of these constants carries a physical meaning

of the motion: A is the amplitude (maximum displacement from the equilibrium position), ω =

2πf is the angular frequency, and φ is the phase.

Using the techniques of differential calculus, the velocity and acceleration as a function of

time can be found (6.7):

𝐝𝐱

𝐝𝐭= 𝐀

𝐝

𝐝𝐭𝐜𝐨𝐬(𝐰𝐭 + 𝛗) = −𝐀𝐰𝐬𝐢𝐧(𝐰𝐭 + 𝛗)

(6.6)

𝐝𝟐𝐱

𝐝𝐭𝟐= −𝐀𝐰

𝐝

𝐝𝐭𝐬𝐢𝐧(𝐰𝐭 + 𝛗) = −𝐰𝟐𝐀𝐜𝐨𝐬(𝐰𝐭 + 𝛗)

(6.7)

If the equation (6.7) write on the number (6.4) equation, can see the equal (6.8).

−𝐰𝟐𝐀𝐜𝐨𝐬(𝐰𝐭 + 𝛗) = −𝐰𝟐𝐀𝐜𝐨𝐬(𝐰𝐭 + 𝛗) (6.8)

Period is the time taken for one full oscillation. Frequency is the number of occurrences of a

repeating event per unit time. The period, usually denoted by T, T =2π

w is the duration of one

cycle, and is the reciprocal of the frequency f, we can expressed the following (6.9) and (6.10);

𝐓 =𝟐𝛑

𝐰= 𝟐𝛑√

𝐦

𝐤

(6.9)

𝐟 =𝟏

𝐓=

𝟏

𝟐𝛑√

𝐤

𝐦

(6.10)

As can be seen, period and frequency depends only on the mass and the spring force constant.

x, is its displacement from the equilibrium (or mean) position, the moving body attached the

spring. A is the amplitude, maximum displacement from the equilibrium position. In simple

harmonic motion displacement (x), velocity (v) and acceleration (a) are varies sinusoidally

with time, but it is not in the same phase. The acceleration (a) of particle is proportional with

their displacement (x) and opposite direction. The frequency of movement is independent of

the amplitude and period.

In addiction; think the moving object, had simple harmonic motion, attached a disc between

two springs as a system spring-disc-spring. The energy of this system is conserved, namely

steady, due to the non- effect forces out of this system. E, total energy of system is equal to

sum of kinetic K, and potential energy U of spring.

𝐊 + 𝐔 = 𝐄 = 𝐒𝐚𝐛𝐢𝐭 (6.11)

where m, the mass of disc , v the velocity of disc, k the constant of the rate or spring constant

and x displacement from the equilibrium, So the kinetic energy of disc is following (6.12);

𝐊 =𝟏

𝟐𝐦𝐯𝟐

(6.12)

and the potential energy of disc is following;

𝐔 =𝟏

𝟐𝐤𝐱𝟐

(6.13)

The total mechanical energy E=K+U, is constant of movement in the harmonic motion, not

effected conservative forces. In the other words the movement is not changed with time. In

this case x displacement is given following;

𝐱 = 𝐀𝐜𝐨𝐬(𝐰𝐭 + 𝛗) (6.14)

According to equation of number (6.13) equation, the values of potential energy are described

at any given moment;

𝐔 =𝟏

𝟐𝐤𝐱𝟐

𝐔 =𝟏

𝟐𝐤𝐀𝟐𝐜𝐨𝐬𝟐(𝐰𝐭 + 𝛗)

(6.15)

The maximum value of potential energy is 1

2kA2 . As shown figure-6.1a and figure-6.1b, the

potential energy is change from zero to maximum value during the movement. The kinetic

energy at any given moment;, 1

2mv2 . the velocity of disc is given the following equations;

𝐯 =𝐝𝐱

𝐝𝐭=

𝐝

𝐝𝐭𝐀𝐜𝐨𝐬(𝐰𝐭 + 𝛗) = −𝐰𝐀𝐬𝐢𝐧(𝐰𝐭 + 𝛗)𝐯𝐞𝐰𝟐 =

𝐤

𝐦

Figure 6.1 a) the kinetic and potential energy of oscillator moving simple harmonic motion, b) the variation of

the kinetic and potential energy according to displacement.

however the kinetic energy of disc are given (6.16);

𝐊 =𝟏

𝟐𝐤𝐀𝟐𝐬𝐢𝐧𝟐(𝐰𝐭 + 𝛗)

(6.16)

As seen, the maximum values of the kinetic energy are found as 1

2kA2 or

1

2m(wA)2 . The

kinetic energy is change from zero to maximum value during the movement. The total energy

for an oscillator is the sum of its kinetic energy and potential energy. The total energy is

obtained by using the equations (6.15) ve (6.16);

𝐄 = 𝐊 + 𝐔 =𝟏

𝟐𝐤𝐀𝟐𝐬𝐢𝐧𝟐(𝐰𝐭 + 𝛗) +

𝟏

𝟐𝐤𝐀𝟐𝐜𝐨𝐬𝟐(𝐰𝐭 + 𝛗)

𝐄 =𝟏

𝟐𝐤𝐀𝟐 (6.17)

As expected, the total mechanical energy is stable and its value is 1

2kA2 . The kinetic energy

is zero and potential energy is 1

2kA2 at the time of the maximum values of displacement. The

potential energy is zero and kinetic energy is 1

2kA2 in the equilibrium position. The total

energy occur the sum of its kinetic energy and potential energy at the other positon.

THE EXPERIMENTAL PROCEDURE

a) To determine the spring constant of a spring

After turning the horizontal air table, attach one of the spring (to be found constant

force) back of the air table and attached the other disc on the free hand of spring.

Place the carbon and measurement paper to the air table.

Press the air pedal intervally and reduce to disc equilibrium position and mark the

position of disc by running the arc chronometer.

After then give different slope the air table (see the Figure-6.2) and Press the air

pedal intervally and reduce to disc equilibrium position and mark the position of

disc by running the arc chronometer every time.

Figure 6.2: Inclined plane, spring-disk system

Do not touch the metal parts of the table during the measurements with spark.

Never switch on the spark if one or two pucks are out of table.

Do not keep one of the pucks at your hand when spark is on.

Otherwise you may get an harmless but disturbing electric shock.

CALCULATIONS

Graphed displacement (x), versus sin(θ) using the result of obtained from the

under the sheet. Describe obtained curves.

Do you say from the shape curve (graph) the constant of spring agree with the

Hook Law within the effect of forces?

If the graph, displacement versus to sin(θ) is linear, write the mathematical

expression. According to, how is relationship between the slope of the line and

k constant spring? Calculate the k constant spring.

b) The spring-disk-spring system and conservation of energy

After turning the horizontal air table, set up the spring-disk-spring system like in

Figure-6.3.

The system must be far away the front of table nearly 10 cm and parallel.

The experiment sheet must fit as keep at both ends passing the carbon sheet

extracting tube.

After then, put the disc side of equilibrium point, press the air pedals and put the

sheet slowly and constant speed.

Thus you will obtain sinusoidal curving as seen Figure-6.4.

This curve is required for proper data depends on the smoothness of paper speed.

Mark the result of projection data on the t and x coordinate.

The path taken along the x-axis of the disk, identify whether equal along the equal

time intervals.

Show the velocity of direction of x Vx on the graph. Determine the values of Vx,

changed the time and find the zero values of Vx on the graph.

Figure 6.3: spring-disk-spring system

Figure 6.4: Displacement with respect to time engaged in simple harmonic motion oscillator.

CALCULATIONS

Find the period T and angular velocity w =2π

T from result of data sheet.

Measure the x values (up to you) of curve part of 1/4.

Use the x values to find potential energy by using U = (1

2)mw2x2 equation.

Likey find the kinetic energy of disc as following equation;

𝐊 =𝟏

𝟐𝐦𝐰𝟐𝐀𝟎

𝟐 −𝟏

𝟐𝐦𝐰𝟐𝐱𝟐 (6.18)

As seen Figure 6.2, find the total energy of system E=K+U for to selected x values.

Plot the graph of total energy versus to x. what’s the slope of obtained curve?

Is curve, showed total energy, parallel?

If it’s parallel what can we say the conservation of total energy?

QUESTIONS

1. Discuss the Hook Law for flexible movements.

2. Calculate the velocity of oscillator (disc) by using figure-4 where point of t =T

4, t =

T

2, t =

3T

4vet = T

3. There is a spring to have vibrated horizontal plane force constant: 20 N/m and mass: 0.5

kg so;

a) If the amplitude of motion A= 3 cm, calculate the total energy of system and maximum

values of velocity of mass.

b) If the displacement x= 2 cm, calculate the kinetic and potential energy of system and

compare the total energy K+U where section a.