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Simple Harmonic Motion vibration Question

  • Thread starter rikiki
  • Start date
  • #1
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Homework Statement


A mass of 0.3kg is suspended from a spring of stiffness 200Nm-1. If the mass is displaced by 10mm from its equilibrium position and released, for resulting vibration, calculate:
a) the frequency of vibration
b) the maximum velocity of the mass during the vibration
c) the maximum acceleration of the mass during the vibration



Homework Equations


Angular frequency= √((Spring constant)/mass)
Frequency= (Angular frequency)/(2 × π)
Velocity: v(t) = -ωA sin(ωt + φ)
Acceleration: a(t) = -ω 2 A cos(ωt + φ)

Displacement = amplitude x sin(angular frequency x time)
T=1/f

The Attempt at a Solution


a) Angular frequency= √((Spring constant)/mass)
ω_n= √(k/m)
ω_n= √(200/0.3)
ω_(n )=25.8199

Frequency= (Angular frequency)/(2 × π)
F= ω_n/2π
F= 25.8199/2π
F=4.11Hz

b) Velocity= -angular frequency ×amplitude ×sin⁡〖(angular frequency ×time + phase constant〗
v(t)= -ωA sin⁡(ωt+φ)
angular frequency=25.8199
amplitude=0.01m
phase constant= ?
time = 0.243


My physics is pretty poor i'm afraid, and i'd be grateful for any help possible. I've become stuck trying to calculate the phase constant. I've read that through using x = A*sin(ωt + φ), and setting t to 0, I can calculate what the phase constant should be. I seem to keep getting 0 as my result here however.Any help on how the phase constant should be calculated would be grately appreciated. Thanks.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,806
932
The "phase constant" is pretty much whatever you want to make it. Specifically, it depends upon what you want to assign as t= 0. However, you first assert that the velocity is given by [itex]-\omega sin(\omega t+ \phi)[/itex] but then you want [itex]x= A sin(\omega t+ \phi)[/itex]. Those can't both be true! The velocity is the derivative of position so if the velocity is given by a sine then x must be a cosine. [itex]x(t)= A cos(\omega t+ \phi)[/itex].

If you choose to make t= 0 at the beginning of the motion, then you must have [itex]Acos(\omega(0)+ \phi)= Acos(\phi)= A[/itex], the maximum motion. Yes, that reduces to [itex]cos(\phi)= 1[/itex] so [itex]\phi= 0[/itex]. What's wrong with that?
 
  • #3
32
0
Thanks for your reply. So I have:

v(t)= -ωA sin⁡(ωt+φ)
angular frequency=25.8199
amplitude=0.01m
time= 1/f=0.2433
phase constant= 0
v(t)= -25.8199 ×0.01 sin⁡(25.8199 × 0.2433+0)
v(t)= -0.258199 sin⁡(6.2817627+0)
v(t)= -0.02825m/s

Is there now further steps involved in calculating the maximum velocity?
 
  • #4
32
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Any body able to offer any more help please.

Where i'm up to:

v(t)= -ωA sin⁡(ωt+φ)
angular frequency=25.8199
amplitude=0.01m
time= 1/f=0.2433
phase constant= 0
v(t)= -25.8199 ×0.01 sin⁡(25.8199 × 0.2433+0)
v(t)= -0.258199 sin⁡(6.2817627+0)
v(t)= -0.02825ms^(-1)

Now i seem to have confused myself and could really do with some help to point me in the right direction.

I need to calculate what the maximum velocity would be. I've read that Maximum and minimum values of any sine and cosine function are +1 and -1. So, would the correct equation be v(t)= -0.258199 sin⁡(1)? Furthermore I've read velocity is at a maximum when displacement is zero. By my understanding this would be when the amplitude is at 0.00. In which case should the equation be v(t)= -25.8199 ×0.00 sin⁡(25.8199 × 0.2433+0).
Both ways seem to give a velocity of 0.00ms?
And is there any other factors that then need adjusting if amplitude is changed? doesn't seem to be, but thought i'd check.

If anybody is able to offer their help that would be great. Thanks.
 
  • #5
Rikiki,

Maximum Velocity is given using

vmax = Aωcos(ωt+∅)

Maximum Acceleration is given using

amax = Aω2sin(ωt+∅)

Other area of this question which I'm struggling with is " (iv) the mass required to produce double the maximum velocity calculated, using the same spring and initial deflection.

Hope this helps a little
 
  • #6
1,506
17
The question asks for max velocity and max acceleration.
Without worrying about phase angle the max velocity occurs as the object passes through the equilibrium position and is given by
v = ωA (A = amplitude)
SWimilarly max acceleration occurs at the max displacement
a = ω^2. A
 
  • #7
im having trouble with;

Plot a graph of acceleration against displacement (x) (for values of x
from x = –10 mm to x = +10 mm)
 
  • #8
10
0
So far I've worked out
f=4.11Hz
vmax=0.26ms-1
amax=0.73ms-2
mass to cause double velocity=0.0755kg

maybe someone can verify these answers.

I'm stumped on this graph though
 
  • #9
10
0
ok forget that, those answers don't work out
f=4.11hz
vmax=0.257ms-1 using either equation above
amax=6.67ms-2 again using either equation

for the graph we can use a= -(2 pi f)*2 xdisplacement to find acceleration at any point during the oscillation.
 
  • #10
1,506
17
In #8 you have max acceleration = 0.73 and in fact it is 6.67 as stated in #9
in SHM acc is proportional to displacement so the graph is a straight line passing through the origin and a = 6.67 when A(amplitude) = +/-10mm
 
Last edited:
  • #11
10
0
thanks technician i got that eventually. for the graph, the value is 6.67 at -10mm straight line passing through zero to -6.67 at +10mm using a=-(2pif)^2 x to find your other acceleration values at displacement points during oscillation
 
  • #12
1
0
Hi

Can anyone please explain how the graph is drawn.

Thanks in advance
 

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