Question

A bob executing simple harmonic motion has an amplitude $A$ and time period $T$. Find the time required by it to travel directly from $y=0$ to $y=\frac{A}{2}$.

• A. $\frac{T}{12}$
• B. $\frac{T}{8}$
• C. $\frac{T}{4}$
• D. $\frac{T}{6}$
• E. None of the above

Answer 1

The correct option is A $\frac{T}{12}$

The general equation of S.H.M. is

$x=ASin\left(\omega t\right)$

where $A$ is the amplitude

$\omega$ is the angular frequency

$\omega =\frac{2\pi }{T}$

From $x=0$ to $x=\frac{A}{2}$

$0=ASin\left(\omega t\right)\Rightarrow \omega t=0\Rightarrow t=0$

$\frac{A}{2}=ASin\left(\omega t\right)\Rightarrow Sin\left(\omega t\right)=\frac{1}{2}\Rightarrow \omega t=Si{n}^{-1}\left(\frac{1}{2}\right)$

$t=\frac{\pi }{6}\times \frac{T}{2\pi }=\frac{T}{12}$

$Timetaken=|\frac{T}{12}-0|=\frac{T}{12}$

For other cases:

From $x=0$ to $x=\frac{A}{\sqrt{2}}$

$0=ASin\left(\omega t\right)\Rightarrow \omega t=0\Rightarrow t=0$

$\frac{A}{\sqrt{2}}=ASin\left(\omega t\right)\Rightarrow Sin\left(\omega t\right)=\frac{1}{\sqrt{2}}\Rightarrow \omega t=Si{n}^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$t=\frac{\pi }{4}\times \frac{T}{2\pi }=\frac{T}{8}$

$Timetaken=|\frac{T}{8}-0|=\frac{T}{8}$

From $x=A$ to $x=\frac{A}{2}$

$A=ASin\left(\omega t\right)\Rightarrow Sin\left(\omega t\right)=1\Rightarrow \omega t=Si{n}^{-1}\left(1\right)$

$t=\frac{\pi }{2}\times \frac{T}{2\pi }=\frac{T}{4}$

$\frac{A}{2}=ASin\left(\omega t\right)\Rightarrow Sin\left(\omega t\right)=\frac{1}{2}\Rightarrow \omega t=Si{n}^{-1}\left(\frac{1}{2}\right)$

$t=\frac{\pi }{6}\times \frac{T}{2\pi }=\frac{T}{12}$

$Timetaken=|\frac{T}{12}-\frac{T}{4}|=\frac{T}{6}$

From $x=\frac{-A}{\sqrt{2}}$ to $x=\frac{A}{\sqrt{2}}$

$\frac{-A}{\sqrt{2}}=ASin\left(\omega t\right)\Rightarrow Sin\left(\omega t\right)=\frac{-1}{\sqrt{2}}\Rightarrow \omega t=Si{n}^{-1}\left(\frac{-1}{\sqrt{2}}\right)$

$t=\frac{5\pi }{4}\times \frac{T}{2\pi }=\frac{5T}{8}$

$\frac{A}{\sqrt{2}}=ASin\left(\omega t\right)\Rightarrow Sin\left(\omega t\right)=\frac{1}{\sqrt{2}}\Rightarrow \omega t=Si{n}^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$t=\frac{\pi }{4}\times \frac{T}{2\pi }=\frac{T}{8}$

$Timetaken=|\frac{T}{8}-\frac{5T}{8}|=\frac{T}{2}$

From $x=\frac{A}{\sqrt{2}}$ to $x=A$

$\frac{A}{\sqrt{2}}=ASin\left(\omega t\right)\Rightarrow Sin\left(\omega t\right)=\frac{1}{\sqrt{2}}\Rightarrow \omega t=Si{n}^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$t=\frac{\pi }{4}\times \frac{T}{2\pi }=\frac{T}{8}$

$A=ASin\left(\omega t\right)\Rightarrow Sin\left(\omega t\right)=1\Rightarrow \omega t=Si{n}^{-1}\left(1\right)$

$t=\frac{\pi }{2}\times \frac{T}{2\pi }=\frac{T}{4}$

$Timetaken=|\frac{T}{4}-\frac{T}{8}|=\frac{T}{8}$