Question

A particle is performing simple harmonic motion having time period $3s$ is in phase with another particle which is also undergoing simple harmonic motion at $t=0$ . The time period of second particle is T (less than 3s). If they are again in the same phase for the third time after $45s$, then the value of T is

• A. $2.8s$
• B. $2.7s$
• C. $2.5s$
• D. $3.2s$

Answer 1

Answer: (C)

$2.5s$

Let ${\omega }_1$ and ${\omega }_2$ be the angular frequencies of first and second particle, respectively. Then, the phase by which they will proceed in time $t$ is ${\omega }_{1}t$ and ${\omega }_{2}t$, respectively.
According to the given situation,
${\omega }_{2}t-{\omega }_{1}t=3\times 2\pi$ for $t=45s$

$\frac{2\pi }{T}-\frac{2\pi }{3}=\frac{3\times 2\pi }{45}$

$\frac{1}{T}=\frac{1}{3}+\frac{1}{15}=\frac{6}{15}\Rightarrow T=2.5s$