Question

A particle is executing $S.H.M$. The time taken for ${\left(\frac{3}{8}\right)}^{th}$ of oscillation from extreme position is $X$ Then the time taken for the particle to complete ${\left(\frac{5}{8}\right)}^{th}$ of oscillation from mean position is

• A. $\frac{5X}{4}$
• B. $\frac{7X}{4}$
• C. $\frac{21X}{8}$
• D. $\frac{7X}{12}$

Answer 1

Answer: (B)

We divide total distance $4A$ in $8$ equal parts.

So, for a displacement of $\frac{3}{8}$ th of an oscillation from an extreme,

Time $=\frac{T}{6}+\frac{T}{12}+\frac{T}{12}=\frac{2+1+1}{12}$
$T=\frac{4}{12}T=\frac{T}{3}$
Given, $\frac{T}{3}=x$ or $T=3x$
Now, for $\frac{5}{8}$ th of oscillation from mean

Time $=\frac{T}{12}+\frac{T}{6}+\frac{T}{6}+\frac{T}{12}+\frac{T}{12}$
$\Rightarrow$ Time for $\frac{5}{8}$ th of oscillations
$=\frac{7}{12}T=\frac{7\times 3x}{12}=\frac{7}{4}x$