Question

A particle of mass \( 1 \, kg \) is hanging from a spring of force constant \( 100 \, Nm^{-1} \). The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is \( \frac{T}{x} \). The value of x is \(\dots\dots\dots\).

Hint:

For any SHM, the energy is shared equally (\( KE = PE \)) when the displacement is \( \frac{1}{\sqrt{2}} \) times the amplitude, which occurs at \( 1/8^{th} \) of the total time period.

Answer 1

Correct Answer: 8

Step 1: Understanding the Concept:
In Simple Harmonic Motion (SHM), the total mechanical energy is conserved and alternates between Kinetic Energy (KE) and Potential Energy (PE). We need to find the specific time \( t \) relative to the period \( T \) when \( KE = PE \).
Step 2: Key Formula or Approach:
1. Displacement equation: \( y = A \sin(\omega t) \).
2. Potential Energy: \( PE = \frac{1}{2} k y^2 \).
3. Kinetic Energy: \( KE = \frac{1}{2} k (A^2 - y^2) \).
Step 3: Detailed Explanation:
Given the condition:
\[ KE = PE \]
\[ \frac{1}{2} k (A^2 - y^2) = \frac{1}{2} k y^2 \]
\[ A^2 - y^2 = y^2 \implies 2y^2 = A^2 \implies y = \frac{A}{\sqrt{2}} \]
Substitute this into the displacement equation (assuming the particle starts from the mean position at \( t=0 \)):
\[ A \sin(\omega t) = \frac{A}{\sqrt{2}} \]
\[ \sin(\omega t) = \frac{1}{\sqrt{2}} \]
The first time this occurs is at:
\[ \omega t = \frac{\pi}{4} \]
Since \( \omega = \frac{2\pi}{T} \):
\[ \left(\frac{2\pi}{T}\right) t = \frac{\pi}{4} \]
\[ t = \frac{T}{8} \]
Comparing this with the given form \( \frac{T}{x} \), we find \( x = 8 \).
Step 4: Final Answer:
The value of x is 8.