Question

A linear harmonic oscillator of force constant \(2\times 10^{6}\) N/m has amplitude \(0.01\) m. Its

• A. maximum potential energy is \(100\) J.
• B. maximum kinetic energy is \(160\) J.
• C. maximum potential energy is \(160\) J.
• D. maximum potential energy is zero.

Hint:

Maximum potential energy at the extreme position is half of the force constant times the square of the amplitude.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a linear harmonic oscillator, the potential energy at displacement \(x\) is \(\frac{1}{2}kx^{2}\). It is maximum at the amplitude \(x=A\).

Step 2: Given Data
\(k=2\times 10^{6}\) N/m and \(A=0.01\) m.

Step 3: Calculation
\(U_{max}=\frac{1}{2}\times 2\times 10^{6}\times(0.01)^{2}=10^{6}\times 10^{-4}=100\) J.

Step 4: Compare the options
The maximum potential energy is 100 J, option (A).

Final Answer:
The maximum potential energy is 100 J. \[ \boxed{\text{(A) }\text{maximum potential energy is }100\text{ J.}} \]