Search any question & find its solution
Question:
Answered & Verified by Expert
A simple harmonic motion is represented by $\alpha \frac{d^2 x}{d t^2}+\beta x=0$. Its period is
Options:
Solution:
1731 Upvotes
Verified Answer
The correct answer is:
$2 \pi \sqrt{\frac{\alpha}{\beta}}$
The standard equation of SHM is: $\frac{d^2 x}{d t^2}=-\omega^2 x$
On comparing with equation: $\alpha \frac{d^2 x}{d t^2}+\beta x=0$
The angular frequency is given by: $\omega=\frac{\beta}{\alpha}$.
Therefore, the time period of SHM is:
$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\alpha}{\beta}}$
On comparing with equation: $\alpha \frac{d^2 x}{d t^2}+\beta x=0$
The angular frequency is given by: $\omega=\frac{\beta}{\alpha}$.
Therefore, the time period of SHM is:
$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\alpha}{\beta}}$
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.