Sound Test

Answers

Question 1

What is the frequency of the pendulum given the graph of its horizontal position as a function of time? Show your work.

Frequency = 1 cycle = .333 Hz 3 seconds

Question 2

What is the period of the pendulum given the graph of its horizontal position as a function of time? Show your work or explain your reasoning.

Period = the time it takes to complete 1 cycle.

It took 3 seconds for this pendulum to complete 1 cycle.

Question 3

What is the length of this pendulum if the acceleration due to gravity is 9.8 m/s2?

Tp = 2 L

g

Question 3

What is the length of this pendulum if the acceleration due to gravity is 9.8 m/s2?

Tp2

= (2 L

g

Question 3

What is the length of this pendulum if the acceleration due to gravity is 9.8 m/s2?

Tp2 g = L

(2

(3.0s)2 9.8 m/s2 = L = 2.2 m

(2

Question 4

a) What would happen to the period and the frequency of oscillation if the acceleration due to gravity changed to 19.6 m/s2? Explain your reasoning.

Question 4

What would happen to the period and the frequency of oscillation if the acceleration due to gravity changed to 19.6 m/s2? Explain your reasoning.

Tp = 2 L g

Question 4

Since the period (Tp) is inversely related to the square root of the acceleration due to gravity (g) the

period will decrease with an increase in

acceleration. The frequency will increase because T = 1/f

Tp = 2 L

g

Question 4

b) What would happen to the frequency if the if the mass of the pendulum bob were

quadrupled? The period of a pendulum is

independent of the mass of the

pendulum and therefore will remain the same.

Question 5

The amplitude of the mass on the string

would be 1.0 m up to the peak

Or

The amplitude of the mass on the spring

would be 1.0 m down to the trough

Question 6

What would happen to the period and frequency of oscillation for this mass spring system if the acceleration due to gravity changed to 19.6 m/s2? Explain your reasoning.

The period and frequency of a mass attached

to a spring are independent of the acceleration due to gravity therefore the period

and frequency remain the same

Question 6

What would happen to the frequency if the mass were quadrupled? Explain your reasoning.

Question 6

What would happen to the frequency if the mass were quadrupled? Explain your reasoning.

Ts = 2 m

K

Question 6

The period would increase by the square

root of 4 or double. The frequency would decrease by the square root of 4 or decrease

by 1/2Ts = 2 m

K

Question 7

What is the period of oscillation of a 1.0 kg mass attached to a 50.0 N / m spring that has been displaced 5.0 cm from its equilibrium position?

Ts = 2 m

K

Question 7

What is the period of oscillation of a 1.0 kg mass attached to a 50.0 N / m spring that has been displaced 5.0 cm from its equilibrium position?

.89 s= Ts = 2 1.0 kg

50 N / m

Question 8

a) A guitar string has linear density of .020 kg / m. The tension on the guitar string is 20.0 N what is the velocity of a wave on the string?

v = FT = 20 N

( m / L) (.020 kg / m)

v = 32 m/s

Question 8

b) If the string with the linear density of .020 kg / m under 20.0 N is .75 meters long what is the lowest frequency that this string can play?

v= f Fundemental L = ½ v = f 2 L

Question 8

b) If the string with the linear density of .020 kg / m under 20.0 N is .75 meters long what is the lowest frequency that this string can play?

v = f = 32 m/s = 21 Hz 2L 2 (.75m)

Question 9

f = 256 Hz= 2 L = 1.60mV = 409.6 m/s

f = 512 Hz =2x256 Hz= L = .80 mV = 409.6 m/s

f = 768 Hz = 3x256 Hz= 2 / 3 L = .53 mV = 409.6 m/s

f = 1024 Hz=4X256 = 1/ 2 L = .40 mV = 409.6 m/s

Question 10

a) Describe or diagram a transverse wave.

Displacement is perpendicular to wave propagation

b) Describe or diagram a longitudinal wave.

Displacement is parallel to wave propagation

Question 10

c) What type of wave is a water wave? transverse d) What type of wave is a sound wave? longitudinal e) What type of wave is a wave on a string? transverse

Question 10

f) v = 331 m/s + .6 T

v= 331 m/s + .6 (20oC) = 343 m/s

Speed of sound is temperature dependent

Question 11 first

11. The following open end tube has a length of .50 m. a) What is the wavelength of this wave form?

Since the tube is .50 m and the wave form represents a whole wave then the wavelength is .50 m

Question 11 first

b) What is the frequency of sound produced by this tube if the speed of sound is 345 m/s?

V = f v = f = 345 m/s = 690 Hz .50 m

Which harmonic does this waveform represent? It represents the 2nd harmonic of a open end

tube

Question 11 first

The following represents the 4th harmonic of a open end tube

Question 11 Second 11. The following closed end tube has a length of .333 m. a) What is the wavelength of this wave form?

This wave represents ¾ of a wave. Therefore L = ¾

or 4/3 L =

m ) = .444 m

Question 11 Second

b) What is the frequency of sound produced by this tube if the speed of sound is 345 m/s?

v = f

v = f = 345 m/s = 777 Hz .444 m

Question 11 Second

Which harmonic does this waveform represent?

This represents the 3rd harmonic of a closed end tube

d) Draw the 1st harmonic or fundamental displacement wave form in the tube below

Question 11 Third

Beat frequency problem 400 Hz and unknown = 20 beats per 10 seconds –

Unknown can be 398 Hz or 402 Hz

401 Hz and unknow =

10 beats per 10 seconds 0

Unknown must be 402 Hz

Question 12

a)Why does a 1000 Hz sound with a sound intensity of 60 dB sound louder than a 100 Hz also with a sound intensity of 60 dB?

We are more sensitive to 1000 Hz sounds than 100 Hz sounds. That is why we need bass boosts in stereo systems.

Question 12

120 db – 80 db = 40 / 10 = 4

dB is a logarithmic scale in which each 10 decibel difference is 10 times more intense sound wave

104 = 10,000 times more energy reaching the ear per meter per second

Question 13

a) Why does the frequency of a sound wave appear to increase when the source and observer move towards each other.

The Doppler effect explains this phenomenon. The relative velocity of the sound increases as the observer and the source approach each other. The number of peaks per second reaching the observer increases and therefore the apparent frequency increases.

Question 14

f = f ’( v + / - vo) ( v - / + vs )

Vo = 30 m/s (observer) vs = -20 m/sThis accounts for there relative velocities toward

each other

f = 500 Hz ( 340 m/s + 30 m/s ) = 578 Hz ( 340 m/s – 20 m/s )

Question 15

a) What is a sonic boom? It is an pressure wave that sounds like thunder b) How are they produced? they are produced

by objects traveling above the speed of sound c) When are they produced? They are produced and released as pressure

builds up on the surface of an object traveling above the speed of sound. This cycle of pressure build up and release repeatedly produce sonic booms

Tacoma Narrows Bridge Breaking Glass Comparison

The Tacoma narrows bridge disaster was caused by standing waves produced in its structure by the wind which caused the bridge to vibrate at its natural frequency. Since the wind was applying a force at the natural frequency of the bridge, constructive interference occurred until the elastic limit of the materials the bridge was composed of was reached.

Tacoma Narrows Bridge Breaking Glass Comparison

Glass can be made to break if the glass is exposed to its natural frequency from a speaker or person at a very high amplitude.

The sound from the person or speaker will cause the beaker to undergo constructive interference ( the glass will resonate ) until the elastic limit of the glass is met and it breaks.