Question

A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximum speed?

Answer 1

Answer:

x=±0.026m

Explanation:

In simple harmonic motion the maximum value of the magnitude of velocity

$v_{max}=wA=\sqrt{\frac{k}{m} }A$

The speed as a function of position for simple harmonic oscillator is given by

$v=w\sqrt{A^{2}-x^{2}$

where A is amplitude of motion

Given data

Amplitude A=3 cm =0.03 m

v=(1/2)Vmax

To find

We have asked to find position x does its speed equal half of is maximum speed

Solution

The speed of the particle the maximum speed as:

$v=\frac{V_{max} }{2}\\ w\sqrt{A^{2}-x^{2} }=\frac{wA}{2}\\ A^{2}-x^{2}=\frac{A^{2} }{4}\\ x^{2}=A^{2}- \frac{A^{2} }{4}\\ x^{2}=\frac{3A^{2}}{4} \\x=\sqrt{\frac{3A^{2}}{4}}$

x=±(√3(0.03)/2)

x=±0.026m