Question:medium

A particle performing linear S.H.M. has period 8 seconds. At time t = 0, it is in the mean position. The ratio of the distances travelled by the particle in the \( 1^{\text{st}} \) and \( 2^{\text{nd}} \) second is (\( \cos 45^\circ = 1/\sqrt{2} \))}

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Distance in the $n^{th}$ interval is $x(n) - x(n-1)$.
Updated On: May 7, 2026
  • \( 1 : (\sqrt{2} - 1) \)
  • \( 1 : 2 \)
  • \( 2 : 1 \)
  • \( 1 : (\sqrt{2} + 1) \)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
In Simple Harmonic Motion (S.H.M.), a particle oscillates about a mean position. The displacement is a sinusoidal function of time.
The distance traveled in a given time interval is the absolute change in position, provided the particle does not reverse its direction within that interval.
Step 2: Key Formula or Approach:
The equation for displacement \( x(t) \) of a particle starting from the mean position at \( t=0 \) is: \[ x(t) = A \sin(\omega t) \] where \( A \) is the amplitude and \( \omega \) is the angular frequency.
Angular frequency is related to the time period \( T \) by: \[ \omega = \frac{2\pi}{T} \] Step 3: Detailed Explanation:
Given the time period \( T = 8 \) seconds.
Calculate the angular frequency \( \omega \): \[ \omega = \frac{2\pi}{8} = \frac{\pi}{4}\text{ rad/s} \] The displacement equation becomes: \[ x(t) = A \sin\left(\frac{\pi}{4} t\right) \] We need to find the distance traveled in the 1st second (from \( t=0 \) to \( t=1 \)) and the 2nd second (from \( t=1 \) to \( t=2 \)).
Note that the particle reaches its extreme position at \( t = T/4 = 8/4 = 2 \) seconds. Thus, it moves in one direction without turning back during the first 2 seconds.
Distance traveled in the 1st second (\( d_1 \)): \[ d_1 = |x(1) - x(0)| = \left| A \sin\left(\frac{\pi}{4} \times 1\right) - A \sin(0) \right| \] \[ d_1 = A \sin\left(\frac{\pi}{4}\right) = \frac{A}{\sqrt{2}} \] Distance traveled in the 2nd second (\( d_2 \)): \[ d_2 = |x(2) - x(1)| = \left| A \sin\left(\frac{\pi}{4} \times 2\right) - A \sin\left(\frac{\pi}{4}\right) \right| \] \[ d_2 = \left| A \sin\left(\frac{\pi}{2}\right) - \frac{A}{\sqrt{2}} \right| \] Since \( \sin(\pi/2) = 1 \): \[ d_2 = A(1) - \frac{A}{\sqrt{2}} = A \left( 1 - \frac{1}{\sqrt{2}} \right) = A \left( \frac{\sqrt{2}-1}{\sqrt{2}} \right) \] Now, calculate the ratio of the distances \( d_1 : d_2 \): \[ \frac{d_1}{d_2} = \frac{\frac{A}{\sqrt{2}}}{A \left( \frac{\sqrt{2}-1}{\sqrt{2}} \right)} \] Cancel \( A \) and \( \frac{1}{\sqrt{2}} \) from numerator and denominator: \[ \frac{d_1}{d_2} = \frac{1}{\sqrt{2} - 1} \] Thus, the ratio is \( 1 : (\sqrt{2} - 1) \).
Step 4: Final Answer:
The ratio of the distances travelled is \( 1 : (\sqrt{2} - 1) \).
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