Question

For a particle in SHM, the displacement $x$ of the particle as a function of time $t$ is given as $x=A\sin (2\pi t)$ Here $x$ is in $cm$ and $t$ is in seconds. Let the time taken by the particle to travel from $x=0$ to $x=A/2$ be $t_1$ and the time taken to travel from $x=A/2$ to $x=A$ be $t_2$. Find $t_1/t_2$

• A. 2/1
• B. 3/1
• C. 1/3
• D. 1/2

Answer 1

Answer: (D)

Here $x=0$ at $t=0$.
Also $\omega =\frac{2\pi }{T}=2\pi \therefore T=1s$
At $t=t_1,x=A/2$. Therefore,
$A/2=A\sin \left(2\pi t_1\right)$ or $1/2=\sin \left(2\pi t_1\right)$
$\therefore 2\pi t_1=\frac{\pi }{6}$ or $t_1=\frac{1}{12}s$
Time taken from $x=0$ to $x=A$ is $\frac{T}{4}=\frac{1}{4}s$
or $t_1+t_2=\frac{T}{4}=\frac{1}{4}s$
or $t_2=\frac{1}{4}-\frac{1}{12}=\frac{1}{6}s$
Hence $\frac{t_1}{t_2}=\frac{1/12}{1/6}=\frac{1}{2}$