Question

In a certain oscillatory system ( particle is performing SHM), the amplitude of motion is $5m$ and the time period is $4s$. The minimum tim etaken by the particle for passing between points, which are at distances of $4m$ and $3m$ from the centre and on the same side of it will approximately be

• A. $\frac{16}{45}s$
• B. $\frac{7}{45}s$
• C. $\frac{8}{45}s$
• D. $\frac{13}{45}s$

Answer 1

Answer: (C)

$\frac{8}{45}s$

$x=A\sin \omega t,$ where $\omega =\frac{2\pi }{4}=\frac{\pi }{2}rad/s$
$3=5\sin \omega t_1\Rightarrow t_1=\frac{1}{\omega }{\sin }^{-1}\left(\frac{3}{5}\right)=\frac{2\times 37}{180}$
$4=5\sin \omega t_2\Rightarrow =\frac{1}{\omega }{\sin }^{-1}\left(\frac{4}{5}\right)=\frac{2\times 53}{180}$
Hence time interval $(t_2-t_1)=\frac{1}{90}(53-37)=\frac{8}{45}s$