Question

A simple harmonic oscillator consists of a particle of mass $m$ and an ideal spring with spring constant $k$. The particle oscillates with a time period $T$. The spring is cut into two equal parts. If one part oscillates with the same particle, the time period will be

• A. 2T
• B. $\sqrt{2}$
• C. $T/\sqrt{2}$
• D. $\frac{T}{2}$

Answer 1

Answer: (C)

$T/\sqrt{2}$

Mass of the particle $=m$

Spring constant $=k$

Time period of the oscillator, $T=2\pi \sqrt{\frac{m}{k}}$

As, $k\propto \frac{1}{l}$ ...... where, $l$ is the length of the spring

$\Rightarrow k^{\prime }=2k$, when the spring is cut into half its length

So, corresponding time period:

$T^{\prime }=2\pi \sqrt{\frac{m}{2k}}=\frac{T}{\sqrt{2}}$