Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2879 Upvotes
Verified Answer
The correct answer is:
$\frac{T}{\sqrt{2}}$
Mass of the particle $=m$
spring constant $=k$
The time period of oscillator, $T=2 \pi \sqrt{\frac{m}{k}}$
As $k \propto \frac{1}{l}$ (where $l$ is the length of spring)
$\begin{array}{rlrl}
\because & k^{\prime} & =2 k \\
& \therefore & T^{\prime} & =2 \pi \sqrt{\frac{m}{2 k}}=\frac{1}{\sqrt{2}} T
\end{array}$
spring constant $=k$
The time period of oscillator, $T=2 \pi \sqrt{\frac{m}{k}}$
As $k \propto \frac{1}{l}$ (where $l$ is the length of spring)
$\begin{array}{rlrl}
\because & k^{\prime} & =2 k \\
& \therefore & T^{\prime} & =2 \pi \sqrt{\frac{m}{2 k}}=\frac{1}{\sqrt{2}} T
\end{array}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.