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displacement ,velocity and acceleration ,phasor diagram questions with solutions
Practice Simple Harmonic Motion for JEE Main, JEE Advanced, NEET, and Classes 11–12. Attempt each question, then open it for the full solution.
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- NumericeasySimple Harmonic Motion
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.
JEE Mains 2024 06 Apr Morning Shift
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A particle executes simple harmonic motion according to x = a sin(50t+π / 3), where x is in centimetres and t is in seconds. Starting from t = 0, let t₁ be the first time at which the particle comes to rest and t₂ be the first time at which its acceleration becomes zero. What are t₁ and t₂, respectively?
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JEE Mains 2026 2 April Evening Shift
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A particle executes SHM of amplitude A along the x-axis. At t = 0, its position is x = A/2 and it moves along the positive x-axis. If x = A sin(ωt + δ), the value of δ is:
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JEE Mains10 April 2023Shift 1Morning Shift
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In SHM x=A sinωt with time period 2 s and amplitude 1 cm, the ratio distance/displacement magnitude during the first 1.25 s is:
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JEE Mains 2025 24 Jan Morning Shift
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Two particles P₁ and P₂ execute SHM about the same mean position with x₁ = A sinωt and x₂ = A sin(ωt+π/3). What is the first positive time at which the two particles occupy the same position?
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JEE MainsNEETJEE Advanced
Open question & full solution → - IntegereasySimple Harmonic Motion
A 5 kg block attached to a horizontal spring executes SHM of amplitude 1 m and time period 3.14 s. The maximum spring force is ______ N.
JEE Mains10 April 2023Shift 2Evening Shift
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A particle executes SHM with amplitude 5 cm. When its displacement from the mean position is 3 cm, its speed is 16 cm/s. If the angular frequency is N rad/s, find N.
JEE MainsNEETJEE Advanced
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Two collinear SHMs of equal amplitude A and equal angular frequency ω have a phase difference of 120°. If they are superposed, what is the amplitude of the resultant SHM?
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JEE MainsNEETJEE Advanced
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A person in an elevator weighs a 50 kg object. The height of the elevator is y = 8[1 + sin(2πt/T)] m, where T = 40π s. Taking g = 10 m s⁻², find the maximum variation of the object's observed weight in newtons.
JEE Advanced 2025 Paper 1
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A particle starts from the mean position and moves towards the positive extreme. Let t₁ be the time required to move from x = 0 to x = A/2, and let t₂ be the time required to move from x = A/2 to x = A. If t₂/t₁ = N, find N.
JEE MainsNEETJEE Advanced
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