Search any question & find its solution
Question:
Answered & Verified by Expert
The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude $=a$) is
Options:
Solution:
2591 Upvotes
Verified Answer
The correct answer is:
$\frac{a}{\sqrt{2}}$
Suppose at displacement $y$ from mean position potential energy = kinetic energy
$\begin{aligned} & \Rightarrow \frac{1}{2} m\left(a^2-y^2\right) \omega^2=\frac{1}{2} m \omega^2 y^2 \\ & \Rightarrow a^2=2 y^2 \Rightarrow y=\frac{a}{\sqrt{2}}\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{1}{2} m\left(a^2-y^2\right) \omega^2=\frac{1}{2} m \omega^2 y^2 \\ & \Rightarrow a^2=2 y^2 \Rightarrow y=\frac{a}{\sqrt{2}}\end{aligned}$
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.