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JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Elasticity & SHM

A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is ______ cm/s.

Answer: 12

Solution

v_{ma x} = A ω = A \frac{2 π}{T} = 0.06 \times \frac{2 π}{3.14} = 0.06 \times 2 = 0.12 m/s = 12 cm/s .

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Integer answer

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