Question

In a certain oscillatory system (particle is performing SHM), the amplitude of motion is $5\text{m}$ and the time period is $4\text{s}$. The minimum time taken (in $\text{s}$) by the particle for passing between points, which are at distance of $4\text{m}$ and $3\text{m}$ from the centre and on the same side of it is $t$. Then value of $45t$ is

Answer 1

Equation of a body starting from mean position and undergoing SHM can be considered as
$x=A\sin \omega t,\text{where}\omega =\frac{2\pi }{4}=\frac{\pi }{2}\text{rad/s}$

Particle reaches $x=3\text{m}$ at time $t_1$
$3=5\sin \omega t_1,\Rightarrow t_1=\frac{1}{\omega }{\sin }^{-1}\left(\frac{3}{5}\right)=\frac{2\times 37}{180}\text{s}$

Particle reaches $x=4\text{m}$ at time $t_2$
$4=5\sin \omega t_2,\Rightarrow t_2=\frac{1}{\omega }{\sin }^{-1}\left(\frac{4}{5}\right)=\frac{2\times 53}{180}$
Hence time interval $(t_2-t_1)=\frac{1}{90}(53-37)=\frac{8}{45}\text{s}$