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An object executes simple harmonic motion with an amplitude \(A\). (a) At what values of its position does its speed equal half its maximum speed? (b) At what values of its position does its potential energy equal half the total energy?

Short Answer

Expert verified
The speed of the object is half its maximum speed at the positions \( ± \frac{A}{\sqrt{2}} \) and the potential energy equals half the total energy at the positions \(±A\).

Step by step solution

01

Compute the position for half maximum speed

First, we equate the speed \(v = \omega \sqrt{A^{2}-x^{2}}\) to half the maximum speed, i.e., \(\frac{\omega \cdot A}{2}\) and solve for \(x\) to get the positions where the speed equals half its maximum speed.
02

Simplify the result

After cancelling \(\omega\) and squaring both sides, we get \( A^{2} - x^{2} = \frac{A^{2}}{4}\) . We solve this equation for \(x\) and simplify the result to obtain \(x = ± \frac{A}{\sqrt{2}}\)
03

Compute the position for half total energy

Now, for part (b), we equate the potential energy \(U = \frac{1}{2} m \omega^{2} x^{2}\) to half the total energy, i.e., \(\frac{1}{4} m \omega^{2} A^{2} \) and solve for \(x\)
04

Simplify the final result

After cancelling \(m\), \(\omega^{2}\), and simplifying the equation, we get \(x = ± A\). This means that the potential energy equals half the total energy at both extremes, i.e., the maximum and minimum position.

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Most popular questions from this chapter

A vertical spring stretches \(3.9 \mathrm{~cm}\) when a \(10-\mathrm{g}\) object is hung from it. The object is replaced with a block of mass \(25 \mathrm{~g}\) that oscillates up and down in simple harmonic motion. Calculate the period of motion.

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