Simple Harmonic Motion of a spring and hammer

In summary, the amplitude of the spring can be calculated using the equation \frac{1}{2}mv_0^2 = \frac{1}{2}kA^2, where m is the mass of the object, v_0 is the initial speed given, and k is the spring constant. The amplitude is the maximum extension of the spring when all of the kinetic energy is converted to potential energy. Using this equation, the amplitude can be determined to be 0.0088m.
  • #1
SnowOwl18
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What seems to be a very simple question is confusing me.

----At t = 0, a 790g mass at rest on the end of a horizontal spring (k = 101N/m) is struck by a hammer, which gives it an initial speed of 2.66m/s. Determine the amplitude.-----

I know that the amplitude is the 'maximum excursion from equilibrium'. I'm not sure how I can calculate that with the given information. In previous problems I calculated the period to be 0.556s and the frequency to be 1.80 Hz. I'm not sure if I need those two values to find it. Could anyone help me out? Thanks :)
 
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  • #2
SnowOwl18 said:
What seems to be a very simple question is confusing me.

----At t = 0, a 790g mass at rest on the end of a horizontal spring (k = 101N/m) is struck by a hammer, which gives it an initial speed of 2.66m/s. Determine the amplitude.-----

You haven't provided all the necessary information. I assume that this is a frictionless surface that the mass is sitting on.

The amplitude is the maximum extension of the spring. This occurs when all of the kinetic energy of the mass is used up and stored as potential energy of the spring. So:

[tex]\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2[/tex]

Solve for A (amplitude).

AM
 
Last edited:
  • #3
that's all the information that was given in the problem. but i just tried your equation and it worked...thanks so much! it was very helpful :)
 

FAQ: Simple Harmonic Motion of a spring and hammer

1. What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion is a type of periodic motion where an object moves back and forth in a straight line with a constant amplitude and a constant frequency. It is caused by a restoring force that is proportional to the displacement of the object from its equilibrium position.

2. How does a spring and hammer demonstrate SHM?

A spring and hammer system can demonstrate SHM when the hammer is attached to the spring and pulled down, causing the spring to stretch. The spring then exerts a restoring force on the hammer, causing it to oscillate up and down in a repetitive motion.

3. What is the relationship between the spring constant and the frequency of SHM?

The spring constant, k, is directly proportional to the frequency of SHM. This means that as the spring constant increases, the frequency of SHM also increases. This relationship can be represented by the formula: f = 1/2π√(k/m), where f is the frequency, k is the spring constant, and m is the mass of the object attached to the spring.

4. What is the relationship between the amplitude and the energy of SHM?

The amplitude of SHM is directly proportional to the energy of the system. This means that as the amplitude increases, the energy also increases. This relationship can be represented by the formula: E = 1/2kA^2, where E is the energy, k is the spring constant, and A is the amplitude.

5. Can the mass of the object attached to the spring affect SHM?

Yes, the mass of the object attached to the spring can affect SHM. A heavier object will have a lower frequency and a longer period of oscillation compared to a lighter object attached to the same spring. This is because the heavier object will have a greater inertia and will require more force to be displaced from its equilibrium position.

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