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Step-by-Step Solution
Step 1: Determine the Angular Frequency and Time Period
Given the mass of the particle, m = 1\,\text{kg} , and the spring constant, k = 100\,\text{N\,m}^{-1} , the angular frequency \omega is found using the formula
\omega = \sqrt{\frac{k}{m}} .
Substituting the values:
\omega = \sqrt{\frac{100}{1}} = 10\,\text{rad/s}.
The time period T of the simple harmonic motion (SHM) is given by
T = \frac{2\pi}{\omega} :
T = \frac{2\pi}{10} = \frac{\pi}{5}.
Step 2: Condition for Kinetic Energy to Equal Potential Energy
In SHM, if x(t) = A\cos(\omega t) is the displacement from the mean (equilibrium) position with amplitude A , then:
Potential Energy, U(t) = \frac{1}{2} k\, x^2 = \frac{1}{2} k \left(A \cos(\omega t)\right)^2.
Kinetic Energy, K(t) = \text{Total Energy} - U(t) = \frac{1}{2} k A^2 - \frac{1}{2} k x^2 = \frac{1}{2} k \bigl(A^2 - x^2\bigr).
For K(t) = U(t) , we set
\frac{1}{2} k\, x^2 = \frac{1}{2} k\, \bigl(A^2 - x^2\bigr).
Canceling \frac{1}{2} k from both sides gives:
x^2 = A^2 - x^2 \quad \Longrightarrow \quad 2\,x^2 = A^2 \quad \Longrightarrow \quad x = \frac{A}{\sqrt{2}}.
Hence, we require the displacement from equilibrium to be \frac{A}{\sqrt{2}}.
Step 3: Find the Required Time at Which x(t) = \frac{A}{\sqrt{2}}
Since x(t) = A\cos(\omega t) , we set
A \cos(\omega t) = \frac{A}{\sqrt{2}} \quad \Longrightarrow \quad \cos(\omega t) = \frac{1}{\sqrt{2}}.
The smallest positive solution is
\omega t = \frac{\pi}{4}.
Thus,
t = \frac{\pi}{4\,\omega}.
Recalling \omega = 10 rad/s,
t = \frac{\pi}{4 \times 10} = \frac{\pi}{40}.
Step 4: Express This Time as a Fraction of T
The time period is T = \frac{\pi}{5} . So,
\frac{t}{T} = \frac{\frac{\pi}{40}}{\frac{\pi}{5}}
= \frac{\pi/40}{\pi/5}
= \frac{1}{8}.
Hence,
t = \frac{T}{8}.
Therefore, if t = \frac{T}{x} and t = \frac{T}{8} , it follows that
x = 8.
Final Answer
The value of x is 8.
