# Simple Harmonic Motion of a spring and hammer

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 :)

Andrew Mason
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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:

$$\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2$$

Solve for A (amplitude).

AM

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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 :)