A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm / s. The distance of the particle from the mean position when its speed becomes 5 cm / s is √α cm, where α= ________.

Solution

Physics sir · 3 steps

Divide the speed at x by the speed at the mean position, and ω drops out on its own.

1. Speed in SHM. At distance x from the mean position, the speed is v = ω√(A² - x²). At the mean position x = 0, so the speed is at its maximum, vₘₐₓ = ω A.

   v/vₘₐₓ = √(A² - x²)/A

2. Put in numbers. Here v/vₘₐₓ = 5/10 = 1/2 and A = 4 cm. So √(16 - x²) = 2, which gives 16 - x² = 4.
3. So. x² = 12, so x = √12 cm and α = 12.

The common slip: 4. Assuming that half the maximum speed means half the amplitude gives x = 2 cm and α = 4. Speed and distance are linked through a square root, not a straight line.

The idea to keep. Speed in SHM does not fall linearly with distance. It follows v = ω√(A² - x²), so half the maximum speed is reached at x = √3/2A, not at A/2.

Answer from the NTA final answer key. Solution by PrepWiser.