Question

A particle in S.H.M. has a period of $4 s$. It takes time $t_{1}$ to start from mean position and reach half the amplitude. In another case it takes a time $t_{2}$ to start from extreme position and reach half the amplitude.

• A. $\frac{t_{1}}{t_{2}}=1$
• B. $\frac{t_{1}}{t_{2}}=\frac{1}{2}$
• C. $\frac{t_{1}}{t_{2}}=2$
• D. $\frac{t_{1}}{t_{2}}=\frac{3}{2}$

Answer 1

Answer: (B)

$y=A \sin (\omega x)$ In the first case $\frac{A}{2}=A \sin \left(\omega t_{1}\right)$
$ \Rightarrow \omega t_{1}=\frac{\pi}{6} \Rightarrow t_{1}=\frac{T}{12}$
In the second case, $\frac{T}{4}-t_{2}=\frac{T}{12} $
$\Rightarrow t_{2}=\frac{T}{6} \& \frac{t_{1}}{t_{2}}=\frac{6}{12}=\frac{1}{2}$