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Simple Harmonic Motion and vibration

  • Thread starter rikiki
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  • #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 maximum velocity 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



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 x 0.01 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.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Doc Al
Mentor
44,892
1,144
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 x 0.01 sin⁡(1)?
Not exactly. Since v = -Aω sin(ωt + φ), the maximum speed will be when sin(ωt + φ) = ± 1. Thus the maximum speed is given by Aω.
Furthermore I've read velocity is at a maximum when displacement is zero.
Right.
By my understanding this would be when the amplitude is at 0.00.
No. The amplitude is a constant of the motion--it doesn't change. The displacement, not the amplitude, is 0.

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?
No. When the displacement is zero, that means cos(ωt + φ) = 0. But velocity, which depends on sin(ωt + φ), is a maximum.
 
  • #3
390
1
Since velocity is given by
[tex]
v = -\omega A\sin(\omega t + \varphi )
[/tex]
And we know that the maximum value of the sin function is 1 the highest value of v must be when [itex] \sin(\omega t + \varphi ) = 1 [/itex] so what does that make the maximum of v? :)

EDIT: Beaten to the punch
 
  • #4
1,506
17
Another nice approach: You know the spring stiffness and the extension therefore you can calculate the FORCE in the stretched spring.
The ELASTIC POTENTIAL ENERGY stored is 0.5 x F x extension
This will be converted to KE of the mass.... use 0.5mv^2 to get maximum v (gives 0.26m/s)
 
  • #5
32
0
That's brilliant, thanks very much for all your help. so I get:

v_max= Amplitude ×angular frequency
v_max = A × ω
v_max=0.01 ×25.8199
v_max=0.258199
v_max=0.258 ms^(-1)

Thanks!
 
  • #6
390
1
I like technician's approach
 

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