Question

A particle performing SHM having time period 3s is in phase with another particle which is also undergoing SHM 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. None

Answer 1

The correct option is C

2.5s

Let ${\omega }_1$ and ${\omega }_2$ be the angular frequencies of $1^{\text{st}}$ and $2^{\text{nd}}$ particle respectively then the phases by which
they will proceed in time $t$ are ${\omega }_{1}t$ and ${\omega }_{2}t$ respectively.
According to given situation
${\omega }_{2}t-{\omega }_{1}t=3\times 2\pi$
For $t=45s$
$\Rightarrow \frac{2\pi }{T}-\frac{1}{3}+\frac{1}{15}=\frac{6}{15}\Rightarrow T=2.5s$