Question

A particle executes SHM with an amplitude of 20 cm and time period of 12 sec. The minimum time required for it to move between two points 10 cm on either side of the mean position

• A. 1 s
• B. 2 s
• C. 3 s
• D. 4 s

Answer 1

Answer: (B)

2 s

Time taken from complete 1 cycle $=t_0+t_0=2t_0$

$T=2t_0$

$\begin{array}{cc}x=A\sin w{t}_0& \\ 10=A\sin \frac{2\pi }{12}{t}_0,& asT=\frac{2\pi }{w},w\frac{2\pi }{T},T=12\\ a=20& \\ \frac{10}{20}=\sin \left(\frac{\pi }{6}{t}_0\right)& \\ \frac{1}{2}=\sin \left(\frac{\pi }{6}{t}_0\right)& \\ \sin \left(\frac{\pi }{6}\right)=\sin \left(\frac{\pi }{6}{t}_0\right)& \\ as\frac{1}{2}=\sin \frac{\pi }{6}& \\ t_0=1& \\ T=2t_0=2\times 1=2\sec .& \end{array}$