Simple Harmonic Motion question

In summary, the problem involves a block being dropped onto a spring and then oscillating with an amplitude of 10 cm. The goal is to find the oscillation frequency. To solve this, the energy equation 1/2k A^2 = 1/2 k x_0 ^2 + 1/2 m v_0 ^2 can be used to find the spring constant k. However, the mass and initial velocity of the block are needed for this equation. Additionally, the block has gravitational potential energy when it is dropped, so this must be taken into account. Another approach is to use the energy at the top of the spring, before the block starts to oscillate, and equate it to the elastic
  • #1
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Homework Statement


A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height 3 cm above the top of the spring. The block sticks to the top end of the spring and then oscillates with an amplitude of 10 cm. What is the oscillation frequency?


Homework Equations



Not sure if this energy equation applies. 1/2k A^2 = 1/2 k x_0 ^2 + 1/2 m v_0 ^2
w = 2 pi / T T = 2pi root( m/ k)
f = 1 /T
w = root ( k/m)

The Attempt at a Solution



I really have no clue on how to start this problem. But i believe i can use the first energy equation to find k . But i am really stuck. Can you guys help me out.
 
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  • #2
You are correct about where to start. You can use the energy equation to find k. Can you do this or are you having trouble computing k? If you are, show me your work and I may be able to help you find your error or help you get past a stumbling block.

After you find k, do you know where to go from there?
 
  • #3
well i don't have the mass for the energy equation, and i am not sure exactly what xo is, is it zero? and what is v_o .
 
  • #4
v_o is whatever the blocks speed was when it hit the spring. Also, you can set x_o as 0 in the problem.

Regarding the the mass of the block, would it help if you solved the equation for k/m instead of just k?
 
  • #5
Thanks for the help.

This is my method, but i am off by .6 from the answer.

i used the kinematics equation to solve for the speed.

Vf ^ 2 = vi ^2 + 2 a s

i get vf = 0.766 m/s

This vf is my new Vi for the spring.

THen i use

1/2 k A^2 = 1/2mvi^2

I solve for k

and i get k = m vi^2 / A^2

I use another formula F = 1/2pi * root ( k/m ) this formula is the frequency formula

I sub k into that formula and i get that the freq is 1.22

but the answer is 1.83. Can someone see if my method is right?
 
  • #6
Your method seems correct to me. Are you sure your numbers are correct?
 
  • #7
There are 2 mistakes I'm seeing here:

i get vf = 0.766 m/s

This vf is my new Vi for the spring.

THen i use

1/2 k A^2 = 1/2mvi^2

first, you're not taking into account the gravitational potential energy of the block... the block is going to drop a certain height...

second the energy stored in the spring is not (1/2)kA^2 when all the energy is converted to elastic potential energy...

The amplitude is the displacement from the equilibrium position. When the mass is at 0 amplitude... there is already energy stored in the spring...

how much has the spring compressed when it is at the equilibrium position... ie: when the mass would be hanging at rest.

so the form of the equation you need is:

1/2 k (A + x0)^2 = 1/2mvi^2 + mg(A + x0).

A little better way, is to just use the energy at the very top (before it moves at all), before it drops to the top of the spring... this way you don't have the kinetic energy term.

(1/2) k (A + x0)^2 = mg(A + x0 + 0.03).

So using the bottom most level the mass reaches as having 0 grav. potential energy, the mass initially has mg(A + x0 + 0.03) energy. At the bottom it only has elastic potential energy in the amount of (1/2) k (A + x0)^2

So find x0... plug it into the above equation... and try to use that equation to find w... try to manipulate the equation so that you won't need to deal with m and k... use the relationship between, w, m and k.
 

1. What is Simple Harmonic Motion (SHM)?

SHM is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium and always directed towards the equilibrium point.

2. What are the key characteristics of SHM?

The key characteristics of SHM include a constant period (time for one complete cycle), constant amplitude (maximum displacement from equilibrium), and the motion follows a sinusoidal pattern.

3. What is the equation for SHM?

The equation for SHM is x = A sin(ωt + φ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

4. What is the difference between SHM and simple pendulum motion?

While both SHM and simple pendulum motion exhibit periodic motion, SHM is characterized by a linear restoring force, while the restoring force of a simple pendulum is due to gravity.

5. How is SHM used in real-life applications?

SHM is used in many real-life applications, such as in the motion of a spring, the swinging of a pendulum, and the vibrations of a guitar string. It is also used in engineering and design of buildings, bridges, and other structures to ensure stability and durability.

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