An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $t = {T \over 4}s$ starting from mean position. Assume that the initial phase of the oscillation is zero.
Solution
$T = 2\pi \sqrt {{m \over k}}$<br><br>$0.2 = 2\pi \sqrt {{{0.5} \over k}}$<br><br>k = 50$\pi$<sup>2</sup><br><br>$\approx$ 500<br><br>x = A sin ($\omega$t + $\phi$)<br><br>= 5 cm sin $\left( {{{\omega T} \over 4} + 0} \right)$<br><br>= 5 cm sin $\left( {{\pi \over 2}} \right)$<br><br>= 5 cm<br><br>$PE = {1 \over 2}k{x^2}$<br><br>$= {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}$<br><br>= 0.6255
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
This question is part of PrepWiser's free JEE Main question bank. 88 more solved questions on Oscillations are available — start with the harder ones if your accuracy is >70%.