PUCET (Punjab) Energy — practice questions

Explore 2 free Energy questions in PUCET (Punjab) Physics. Open each question or exercise to read its content, reference answers and available explanation.

Practise Energy with Quill →

Energy questions and exercises with answers

Showing questions 1–2 of 2.

Total mechanical energy of a particle in SHM (mass $m$, amplitude $A$, angular frequency $\omega$) is:Read the answer and worked solution →In SHM, when the displacement is half the amplitude (x = A/2), the ratio of KE to PE is:Read the answer and worked solution →

Worked practice questions

Total mechanical energy of a particle in SHM (mass $m$, amplitude $A$, angular frequency $\omega$) is:

$\frac{1}{2} m \omega A$
$\frac{1}{2} m \omega^2 A^2$
$m \omega^2 A^2$
$\frac{1}{2} m \omega A^2$

Show answer and solution

Correct answer: B. $\frac{1}{2} m \omega^2 A^2$

$E = \frac{1}{2} m \omega^2 A^2$. Constant throughout the motion. At any point, $E = KE + PE = \frac{1}{2}m\omega^2(A^2 - x^2) + \frac{1}{2}m\omega^2 x^2$.

Open this question →

In SHM, when the displacement is half the amplitude (x = A/2), the ratio of KE to PE is:

1 : 1
2 : 1
3 : 1
4 : 1

Show answer and solution

Correct answer: C. 3 : 1

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$ $PE = \frac{1}{2}m\omega^2 x^2$. At $x = A/2$: $KE \propto A^2 - A^2/4 = 3A^2/4$ $PE \propto A^2/4$. Ratio = 3:1.