Question

A particle is executing SHM of amplitude \(4\text{ cm}\) and time period \(4\text{ s}\). The time taken by it to move from positive extreme position to half the amplitude is

• A. \(1\text{ s}\)
• B. \(\frac{1}{3}\text{ s}\)
• C. \(\frac{2}{3}\text{ s}\)
• D. \(\sqrt{\frac{2}{3}}\text{ s}\)

Hint:

Standard SHM time intervals to memorize:
• Mean position (\(0\)) to half amplitude (\(A/2\)): \(t = T/12\)
• Half amplitude (\(A/2\)) to extreme (\(A\)): \(t = T/6\)
• Extreme (\(A\)) to half amplitude (\(A/2\)): \(t = T/6\) Here, the particle moves from \(A\) to \(A/2\), so \(t = T/6 = 4/6 = 2/3\text{ s}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Simple Harmonic Motion (SHM) describes the oscillatory motion of a particle where the restoring force is directly proportional to displacement.
The displacement \( x \) of such a particle at any time \( t \) can be represented using sinusoidal functions.
When a particle starts its motion from the mean position \( (x=0) \), the sine function is typically used: \( x = A \sin(\omega t) \).
When the particle starts from the positive extreme position \( (x=A) \), it is more convenient to use the cosine function: \( x = A \cos(\omega t) \).
In this specific problem, the particle moves from the extreme position, so we proceed with the cosine equation.
The speed of the particle is zero at the extremes and maximum at the mean position.
Step 2: Key Formula or Approach:
The equation for displacement from the extreme position is:
\[ x = A \cos(\omega t) \]
Where:
\( A = \) Amplitude of oscillation.
\( \omega = \) Angular frequency, defined as \( \omega = \frac{2\pi}{T} \).
\( T = \) Time period of the oscillation.
Given in the problem: \( A = 4 \) cm and \( T = 4 \) s.
Step 3: Detailed Explanation:
First, calculate the angular frequency \( \omega \):
\[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/s} \]
The problem asks for the time taken to reach "half the amplitude".
This means the final displacement \( x \) should be:
\[ x = \frac{A}{2} = \frac{4}{2} = 2 \text{ cm} \]
Substitute the values of \( x, A, \) and \( \omega \) into the displacement equation:
\[ 2 = 4 \cos\left( \frac{\pi}{2} t \right) \]
Divide both sides by 4 to isolate the cosine term:
\[ \frac{2}{4} = \cos\left( \frac{\pi}{2} t \right) \]
\[ \frac{1}{2} = \cos\left( \frac{\pi}{2} t \right) \]
From trigonometric tables, we know that \( \cos \theta = \frac{1}{2} \) when \( \theta = \frac{\pi}{3} \) radians.
Equate the arguments of the cosine function:
\[ \frac{\pi}{2} t = \frac{\pi}{3} \]
Cancel \( \pi \) from both sides:
\[ \frac{t}{2} = \frac{1}{3} \]
Multiply by 2 to solve for \( t \):
\[ t = \frac{2}{3} \text{ seconds} \]
This represents the time elapsed for the particle to travel from \( x=4 \) cm to \( x=2 \) cm.
The motion of the particle in SHM is non-uniform; it takes more time to cover distances near the extremes compared to the same distances near the mean position.
Step 4: Final Answer:
The time taken by the particle to move from the positive extreme to half the amplitude is \( \frac{2}{3} \) s.
This corresponds to Option (C).