MHT CET · Physics · Oscillations
Ratio of kinetic energy at mean position to potential energy at \(A / 2\) of a partide performing SHM
- A \(2: 1\)
- B \(4: 1\)
- C \(8: 1\)
- D \(1: 1\)
Answer & Solution
Correct Answer
(B) \(4: 1\)
Step-by-step Solution
Detailed explanation
Kinetic energy \(K=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)\)
At mean position \(y=0\)
\(
K=\frac{1}{2} m \omega^{2}\left(A^{2}\right)
\)
Potential energy \(U=\frac{1}{2} m \omega^{2} y^{2}\)
\(U\) at \(y=\frac{A}{2}\)
\(
U=\frac{1}{2} \frac{m A^{2}}{4} \omega^{2}
\)
Dividing Eq (i) by Eq (ii), we get
\(
\frac{\mathrm{KE}}{U}=\frac{4}{1}
\)
At mean position \(y=0\)
\(
K=\frac{1}{2} m \omega^{2}\left(A^{2}\right)
\)
Potential energy \(U=\frac{1}{2} m \omega^{2} y^{2}\)
\(U\) at \(y=\frac{A}{2}\)
\(
U=\frac{1}{2} \frac{m A^{2}}{4} \omega^{2}
\)
Dividing Eq (i) by Eq (ii), we get
\(
\frac{\mathrm{KE}}{U}=\frac{4}{1}
\)
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