MHT CET · Physics · Oscillations
The total energy of the body executing simple harmonic motion is \(E\). When the displacement is half of the amplitude then the kinetic energy is
- A \(\frac{E}{4}\)
- B \(\frac{3 E}{4}\)
- C \(\frac{\sqrt{3} E}{4}\)
- D \(\frac{E}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{3 E}{4}\)
Step-by-step Solution
Detailed explanation
Total energy is SHM \(E=\frac{1}{2} m \omega^2 a^2\) (where \(a=\) amplitude)
The potential energy at displacement \(y\) is \(U=\frac{1}{2} m \omega^2 y^2\)
The kinetic energy is \(K=E-U=\frac{1}{2} m \omega^2\left(a^2-y^2\right)=E-\frac{1}{2} m \omega^2 y^2\)
Now at \(y=\frac{a}{2}\)
\(K=E-\frac{1}{2} m \omega^2\left(\frac{a^2}{4}\right)=E-\frac{E}{4}=\frac{3 E}{4}\)
The potential energy at displacement \(y\) is \(U=\frac{1}{2} m \omega^2 y^2\)
The kinetic energy is \(K=E-U=\frac{1}{2} m \omega^2\left(a^2-y^2\right)=E-\frac{1}{2} m \omega^2 y^2\)
Now at \(y=\frac{a}{2}\)
\(K=E-\frac{1}{2} m \omega^2\left(\frac{a^2}{4}\right)=E-\frac{E}{4}=\frac{3 E}{4}\)
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