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JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Elasticity & SHM

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of

4 m, 2 ms^{- 1}

and

16 ms^{- 2}

at a certain instant. The amplitude of the motion is

x m

where x is ____.

Answer: 17

Solution

∣ a ∣ = ω^{2} x \Rightarrow 16 = ω^{2} (4) \Rightarrow ω = 2

rad/s.

v = ω A^{2} - x^{2} \Rightarrow A = \frac{v ^{2}}{ω ^{2}} + x^{2} = \frac{4}{4} + 16 = 17

m, so

x = 17

.

JEE Main · 2024 · 09 Apr 2024 · Shift 1 · Integer answer

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