A particle is executing SHM. The time taken for $\left(\frac{3}{8} ight)^{ ext {th }}$ of oscillation from extreme positions is $x$. Then, the time taken for the particle to complete $\left(\frac{5}{8} ight)^{ ext {th }}$ of oscillation from mean position is

Question

A particle is executing SHM. The time taken for $\left(\frac{3}{8}ight)^{	ext {th }}$ of oscillation from extreme positions is $x$. Then, the time taken for the particle to complete $\left(\frac{5}{8}ight)^{	ext {th }}$ of oscillation from mean position is

Aicharya · Accepted Answer

We divide total distance $4 A$ 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 $T=3 x$ Now, for $\frac{5}{8}$ th of oscillation from mean

$ \text { 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 3 x}{12}=\frac{7}{4} x $

A particle is executing SHM. The time taken for $\left(\frac{3}{8}\right)^{\text {th }}$ of oscillation from…

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