SHM mass and spring constant with given frequency

In summary, the conversation discusses a problem involving a block-spring system with a given frequency and an added mass that causes a decrease in frequency. The conversation goes on to discuss the use of equations to solve for the mass and spring constant. The attempted solution is shown, including the steps to find the angular frequency for each situation and setting up the equations for the unknowns. The error is found to be in the step where the equations are set equal and the correct values for the mass and spring constant are then provided.
  • #1
k77i
28
0

Homework Statement



With a block of mass m, the frequency of a block-spring system is 1.5Hz. When 80g is added, the frequency drops to 0.9Hz. Find m and the spring constant k.


Homework Equations



T=1/f
w=2pi/T
so w= 2pi*f
w^2=k/m
so k=mw^2
T=2pi*sqrt(m/k)

The Attempt at a Solution



The problem seems simple enough. I just have to set up 2 equations to solve one unknown and then find the other right?
So first i found the w for when the mass= m and f= 1.5Hz (i called this w) and the w for when mass= m+0.08 and f= 0.9Hz (W).

Using the formula w= 2pi*f, i found 9.42 for w and 5.65 for W.

Then I set up the 2 equations:

k = mw^2 = 88.8m
and
k = (m+0.08)W^2 = 31.9m + 2.55

After i did equation 1 = equation 2, so:
88.8m = 31.9m + 2.55 and this gave me m=22.3

But the correct answer for m is supposed to be 0.045 kg. Can somebody let me know what I'm doing wrong please?

The correct value for k is supposed to be 3.99N/m
 
Physics news on Phys.org
  • #2
k77i said:
88.8m = 31.9m + 2.55 and this gave me m=22.3

Try redoing the above step again. I think that's where the trouble is coming from.
 
  • #3
I'm not exactly sure which steps you're asking me to redo but i'll show my calculations in more detail if that's what you meant.

w= 2pi*f1 = 2pi*1.5 = 9.42 rad/s
W= 2pi*f2 = 2pi*0.9 = 5.65 rad/s

Then,

1) k= mw^2 = m*9.42^2 = 88.8m
2) k= W^2(m+0.08) = 5.65^2(m+0.08) = 31.9m + 2.55

I really still don't see what's wrong with the way I'm doing it..
 
  • #4
k77i said:
I'm not exactly sure which steps you're asking me to redo but i'll show my calculations in more detail if that's what you meant.

w= 2pi*f1 = 2pi*1.5 = 9.42 rad/s
W= 2pi*f2 = 2pi*0.9 = 5.65 rad/s

Then,

1) k= mw^2 = m*9.42^2 = 88.8m
2) k= W^2(m+0.08) = 5.65^2(m+0.08) = 31.9m + 2.55

I really still don't see what's wrong with the way I'm doing it..

So far so good. It's the next step where you solve for m (after you set the equations equal).
 
  • #5
Well ok..

Equation 1 = Equation 2

88.8m = 31.9m + 2.55

then bring the "m" terms to one side

88.8m - 31.9m = 2.55

56.6m = 2.55

and then divide..

Ahhh! wow i feel really stupid.. i was dividing both sides by 2.55 all this time.. I get it now, thanks a lot
 

Related to SHM mass and spring constant with given frequency

1. What is SHM?

SHM stands for Simple Harmonic Motion, which is a type of motion where the restoring force is directly proportional to the displacement from equilibrium and is in the opposite direction of the displacement.

2. How is mass related to SHM?

Mass does not affect the frequency of SHM, but it does affect the amplitude of the oscillations. A larger mass will result in smaller amplitudes, while a smaller mass will result in larger amplitudes.

3. What is the spring constant in SHM?

The spring constant, also known as the force constant, is a measure of the stiffness of the spring. It is represented by the letter k and is directly proportional to the frequency of SHM.

4. How do you calculate the frequency of SHM with given mass and spring constant?

The formula for the frequency of SHM is f = 1/2π √(k/m), where k is the spring constant and m is the mass. Simply plug in the values for k and m to calculate the frequency.

5. Can the frequency of SHM be changed by adjusting the mass or spring constant?

Yes, the frequency of SHM can be changed by adjusting either the mass or the spring constant. Increasing the mass will decrease the frequency, while increasing the spring constant will increase the frequency.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
383
  • Introductory Physics Homework Help
Replies
8
Views
390
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
12
Views
288
  • Introductory Physics Homework Help
Replies
3
Views
402
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
17
Views
495
  • Introductory Physics Homework Help
Replies
7
Views
6K
Back
Top