Do now: Thursday, April 20, 2023
A mass oscillates with SHM, period T = 2.4 s and amplitude A = 0.12m
1) Use trig to calculate the angle
2) Calculate the time taken to travel from point C to B
3) Calculate the time taken to travel from point B to A
Quickly try: Thursday, April 20, 2023
A mass oscillates with SHM, period T = 2.4 s and amplitude A = 0.12m
4) Where is the mass at t = 1.0s?
5) What is the velocity of the mass at t = 1.0s?
6) What is the acceleration of the mass at t = 1.0s?
More practise: Thursday, April 20, 2023
A mass oscillates with SHM, period T = 2.4 s and amplitude A = 0.12m
7) Where is the mass at t = 1.8s?
8) What is the velocity of the mass at t = 1.8s?
9) What is the acceleration of the mass at t = 1.8s?
44
SHM VelocitySHM Velocity
For For pp and and mm to be to be staying in line…staying in line…
……they must both they must both have the same have the same vertical component vertical component to their speeds at to their speeds at any time.any time.
pm
A
vm vp
What is the relationship between the What is the relationship between the SHM velocity SHM velocity vvmm and the velocity of the and the velocity of the test particle test particle vvpp on the reference circle? on the reference circle?
So substituting this back into the first So substituting this back into the first equation:equation:
SHM VelocitySHM Velocity vvmm must be equal must be equal
to the vertical to the vertical component of component of vvpp::
vmvp
cosmv Aω ωt
cosm pv v The reference circle has The reference circle has
constant constant vvpp::
pv r A
This formula will be given in the exam. Extension: derive using calculusThis formula will be given in the exam. Extension: derive using calculus
66
Displacement-timeDisplacement-time A graph of displacement-time can be A graph of displacement-time can be
drawn by using a rotating radius drawn by using a rotating radius vector known as a vector known as a Phasor.Phasor.
tω t=0
t=1
t=2
t=3
t=5
t=6
t=7
1 72 3 54 6 8
77
Sketch the velocity-time graph Sketch the velocity-time graph for SHMfor SHM
tω t=6
t=7
t=0
t=1
t=3
t=4
t=5
1 72 3 54 6 8
+Aω
– Aω
VelocityWatch video clip
88
Velocity PhasorVelocity Phasor
tω
1 72 3 54 6 8
velocity
displacement
• The length of the velocity phasor is…The length of the velocity phasor is…• What is the phase difference between the velocity phasor What is the phase difference between the velocity phasor
and the diplacement phasor? and the diplacement phasor?
The velocity phasor is 90° (or The velocity phasor is 90° (or /2 rad) /2 rad) aheadahead of the of the displacement phasor displacement phasor
AA
99
SHM AccelerationSHM Acceleration Point Point pp has centripetal has centripetal
acceleration acceleration aapp in in towards the centre of towards the centre of the circlethe circle
The SHM object has The SHM object has acceleration acceleration aam m
towards equilibrium...towards equilibrium...
...which must be ...which must be equal to the vertical equal to the vertical component of component of aapp
pm
am ap
1010
SHM AccelerationSHM Acceleration
am
ap
sinm pa a
Negative sign indicates direction is always
towards equilibrium
2 sinma A t
ButBut a app is centripetal, so: is centripetal, so:
2
pp
va
r 2
A
A
2A
Putting this all together:Putting this all together:
1111
t
t=5
ωt=0
t=6
t=7
t=1
t=2
t=3
1 72 3 54 6 8
Sketch the acceleration-time Sketch the acceleration-time graph for SHMgraph for SHM
1212
All three….All three….
tω
1 72 3 54 6 8
velocity
displacement
acceleration
Clip: Spring SHM with d v and a graphs alongside - 20sec
1313
All three…All three…
If t=0 is not equilibrium, then all the phasors If t=0 is not equilibrium, then all the phasors shift their starting positions.shift their starting positions. The graphs start The graphs start in different positions and the sin/cosine in different positions and the sin/cosine function may change.function may change.
tAa
tAv
tAy
sin
cos
sin
2
Do pg 124 Qu 13-22
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