Distance Travelled by Particle in SHM – Rankers Physics
Topic: Oscillation
Subtopic: Equation of SHM

Distance Travelled by Particle in SHM

A particle is performing simple harmonic motion of amplitude A and time period 12 seconds. If at t = 0 particle is at mean position, then distance travelled by particle in first 5 seconds will be:
\(\frac{7A}{3}\)
\(\frac{4A}{3}\)
\(\frac{3A}{2}\)
\(\frac{5A}{2}\)

Solution:

The equation of motion is \(x = A \sin\left(\frac{\pi t}{6}\right)\). At \(t = 3\text{ s}\), \(x = A\). At \(t = 5\text{ s}\), \(x = A \sin(5\pi/6) = 0.5A\). Total distance is \(A\) (from 0 to \(A\)) plus \(0.5A\) (returning from \(A\) to \(0.5A\)), which is \(1.5A = \frac{3A}{2}\).

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