Question

A particle executes $S.H.M$ with amplitude $‘a’$ and time period $T$. The displacement of the particle from the mean position when its speed is half of the maximum speed is $\surd xa/2$. The value of $x$ is ________.

Answer 1

Step 1: Given data

Amplitude $=a$

Time period $=T$

Displacement $=\surd xa/2$

$V=V_{max}/2\Rightarrow a\omega /2$

Step 2: Find value of $x$

For a particle executing $S.H.M$

$$\begin{gathered} V=\omega \surd (a^2–x^2) \\ {a}^2{\omega }^2/4={\omega }^2{a}^2–{\omega }^2{x}^2 \\ x=(\surd 3/2)a \end{gathered}$$

Hence, The value of $x$ is $3$.