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 $\omega=\frac{2 \pi}{T}=2 \pi \quad \therefore T=1 s$
At $t=t_{1}, x=A / 2$. Therefore,
$A / 2=A \sin \left(2 \pi t_{1}\right)$ or $1 / 2=\sin \left(2 \pi t_{1}\right)$
$\therefore 2 \pi t_{1}=\frac{\pi}{6} \quad$ 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 $\frac{t_{1}}{t_{2}}=\frac{1 / 12}{1 / 6}=\frac{1}{2}$