Question

The variation in potential energy of a harmonic oscillator is as shown in the figure. The spring constant is


A. \[1 \times {10^2}N{m^{ - 1}}\]
B. \[1.5 \times {10^2}N{m^{ - 1}}\]
C. \[0.0667 \times {10^2}N{m^{ - 1}}\]
D. \[3 \times {10^2}N{m^{ - 1}}\]

Answer 1

Hint: The term potential energy refers to the energy that has been stored or the energy that has been caused by its location. The potential energy of an object changes mostly when the object is lower or higher than its original position.

Formula Used:
\[PE = {V_0} + \dfrac{1}{2}k{x^2}\]
where PE is the potential energy, \[{V_0}\] is the initial energy, k is the spring constant and x is the displacement travelled by the string.

Complete step by step solution:
We have been given that the initial energy of the harmonic oscillator is 0.01 J and its final potential energy is 0.04 J. The spring of the harmonic oscillator has travelled 20 mm or 0.02 m. We have to find the spring constant of the oscillator. Let \[{V_0}\] be the initial potential energy, PE be the final potential energy, x be the distance travelled by the string, and k be the spring constant.

By using the formula of the potential energy of the harmonic oscillator, we get,
\[PE = {V_0} + \dfrac{1}{2}k{x^2} \\
\Rightarrow 0.04 = 0.01 + \dfrac{1}{2}k{\left( {0.02} \right)^2} \\
\Rightarrow 0.03 = 0.0002k \\
\Rightarrow k = 150N{m^{ - 1}} \]

Writing the result in the scientific notation we get,
\[k = 150N{m^{ - 1}} \\
\Rightarrow k = 1.5 \times 100N{m^{ - 1}} \\
\therefore k = 1.5 \times {10^2}N{m^{ - 1}} \]

So, option B, \[1.5 \times {10^2}N{m^{ - 1}}\] is the required solution.

Note: As the harmonic oscillator initially had some potential energy and after applying the spring its potential energy gets increased so to calculate the spring constant we only need the energy generated by the spring. So, for the energy generated only by spring, we subtract the initial potential energy from the final potential energy.