Calculating Frequency in SHM with Amplitude, Velocity, and Mass

  • Thread starter Thread starter izzyfish
  • Start date Start date
  • Tags Tags
    Shm
Click For Summary
To calculate frequency in simple harmonic motion (SHM) using amplitude, velocity, and mass, the relationship between these variables is key. The participant attempted to derive frequency using the equations for position and velocity in SHM, ultimately calculating a frequency of approximately 0.0795 Hz. They also explored using Hooke's Law to find the spring constant and subsequently the frequency, yielding a slightly different result of 0.1576 Hz. There is confusion regarding the correct method to use, particularly in relation to the orientation of the spring and the applicability of the equations. Clarification on the proper approach to derive frequency from the given parameters is sought.
izzyfish
Messages
1
Reaction score
0
I feel like an idiot for not being able to answer this question, it seemed so simple yet so complicated in my head.

Homework Statement


Have to find frequency, given the amplitude, velocity, and mass.
I don't have numbers as it was off an exam I just took. Guessing it was v=5 m/s; A=10m; and m=3kg.


Homework Equations


I was thinking of taking the derivative of the SHM eq x(t) = Asin(wt) but then I wasn't sure if I could calculate the spring constant with F = -K(xo - xf).


The Attempt at a Solution


My attempt was guessing, but if we threw in numbers, A=10m, v=5m/s, m=3kg...
x(t) = A * sin(wt)
w = 2 * PI * f
x(t) = A * sin(2 * PI * f * t)
v(t) = 2 * PI * f * A * cos(2 * PI * f * t)
t = 0
5 = 2 * PI * f * 10 * cos(2 * PI * f * 0)
5 = 2 * PI * f * 10 * 1
5 = 20 * PI * f
5 / (20 * PI) = f
f = 0.0795Hz

or..

F=-K(x-xo)
F = 3*9.8 = 29.4
x = 10
xo = 0
K = 2.94 N/m
f = (1 /( 2 * PI )) * sqrt( k / m)
f = 0.159 * sqrt (2.94 / 3 )
f = 0.1576 Hz

Would love to know the correct way to do this, as I'm confused..
 
Physics news on Phys.org
Using F=kx would depend on how the spring is orientated. But I would use your first method.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
850
Replies
3
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
824
Replies
5
Views
745
  • · Replies 10 ·
Replies
10
Views
2K
Replies
39
Views
4K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
4
Views
992