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## Homework Statement

Consider a block-spring system that forms a linear simple harmonic oscillator, with the block undergoing SHM. The angular frequency is 9.78 rad/s, the frequency is 1.6 Hz and the period is 640 msec. The block is pulled from its equilibrium position at x=0 (at time = 0) to x=11 cm. It is released from x=11cm. What is the amplitude.

## Homework Equations

I tried using x(t) = A[cos(ωt + ∅)]

## The Attempt at a Solution

Ok, so I know that the answer this question is intuitively 11 cm, since the block can never go further than 11 cm from the equilibrium point. However, I thought I'd try testing my shaky understanding of the x(t) = A[cos(ωt + ∅)] formula and I hit a brick wall.

I figured that it would take one quarter of a period for the block to get from x=11 cm to x = 0 cm, since that's one fourth of the journey through the period. So, I plugged the following numbers in:

.11 m = (A)cos[(9.78rad/s)(.643sec/4) + 0]

A = -82.17 m

I don't understand why this method doesn't yield A = 11 cm (or anything remotely close to it).

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