Question

A particle executes simple harmonic motion with an amplitude of 2.7 cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its maximum speed? Answer in units of cm.

Answer 1

Given,

Amplitude (A)=2.7 cm

Maximum speed (V)=$\frac{V_{max}}{2}$

$V=\omega \sqrt{a^2-x^2}$

Now, substituting the values,

$\frac{V_{max}}{2}=\omega a\sqrt{1-\frac{x^2}{a^2}}$

$\Rightarrow \frac{V_{max}}{2}=V_{max}\sqrt{1-\frac{x^2}{a^2}}$

$\Rightarrow \frac{1}{2}=\sqrt{1-\frac{x^2}{a^2}}$

Now, squaring both side,

$\Rightarrow \frac{1}{4}=1-\frac{x^2}{a^2}$

Now, substituting the values,

$\Rightarrow \frac{1}{4}=1+\frac{x^2}{2.7^2}$

$\Rightarrow \frac{x^2}{2.7^2}=\frac{3}{4}$

$\Rightarrow x^2=\frac{3}{4}\times 2.7^2$

$\Rightarrow x = 2.7\times \frac{\sqrt{3}}{2}cm$