Question

A linear harmonic oscillator of force constant $ 2 \times 10^6 $ N/m and amplitude 0.01 m has a total mechanical energy of 160 J. Show that its (a) maximum potential energy is 160 J (b) maximum kinetic energy is 100J

• A. P.E. is 160 J
• B. P.E. is zero
• C. P.E. is 100 J
• D. P.E. is 120 J

Answer 1

The Correct Option is C

The problem involves a linear harmonic oscillator, and we need to determine its maximum potential energy and maximum kinetic energy. We are given the force constant, amplitude, and total mechanical energy. Let's analyze the problem step-by-step:

Given Data:

• Force constant k = 2 \times 10^6 N/m
• Amplitude A = 0.01 m
• Total mechanical energy E = 160 J

Step-by-Step Solution:

(a) Maximum Potential Energy

The potential energy (P.E.) in a harmonic oscillator at maximum displacement (amplitude) is given by the formula:

\text{P.E.} = \frac{1}{2} k A^2

Substituting the given values:

\text{P.E.} = \frac{1}{2} \times 2 \times 10^6 \, \text{N/m} \times (0.01 \, \text{m})^2

\text{P.E.} = \frac{1}{2} \times 2 \times 10^6 \times 0.0001

\text{P.E.} = 100 \, \text{J}

So, the maximum potential energy is 100 J.

(b) Maximum Kinetic Energy

The total mechanical energy is the sum of the maximum potential energy and maximum kinetic energy. Therefore, the maximum kinetic energy (K.E.) is:

\text{K.E.} = E - \text{P.E.}_{\text{max}}

\text{K.E.} = 160 \, \text{J} - 100 \, \text{J}

\text{K.E.} = 60 \, \text{J}

So, the maximum kinetic energy is 60 J.

Conclusion:

The maximum potential energy of the linear harmonic oscillator is 100 J, and the maximum kinetic energy is 60 J. This is a verification problem where we use the formulas for potential and kinetic energy in harmonic oscillators to confirm the values given in the question, but it seems there's an error mentioned in the question regarding kinetic energy being 100 J.