alleryStart learning →Maximum acceleration of a particle in SHM (amplitude A, angular frequency ω):
- A²ω
- A
- Aω²
- Aω
Show answer and solution
Correct answer: C. Aω²
a_max = Aω², occurring at extreme positions (x = ±A).
From a = -ω²x, max |a| = ω²A.
AMUEEE Oscillations — practice questions — page 2
Explore 75 free Oscillations questions in AMUEEE Physics. Open each question or exercise to read its content, reference answers and available explanation.
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Showing questions 31–60 of 75.
Maximum acceleration of a particle in SHM (amplitude A, angular frequency ω):Read the answer and worked solution →A particle executing SHM has time period T. Time taken to go from extreme position (x = A) to x = A/2:Read the answer and worked solution →A simple pendulum of length 1 m has period T (g = 9.8). New length needed to double T:Read the answer and worked solution →Total energy of SHM particle equals:Read the answer and worked solution →At the extreme position in SHM, kinetic energy equals:Read the answer and worked solution →Two springs k1 and k2 in parallel: equivalent spring constant:Read the answer and worked solution →Two springs k1 and k2 in series: equivalent spring constant:Read the answer and worked solution →A spring of force constant k is cut into two equal halves. Spring constant of each half:Read the answer and worked solution →A particle executing SHM has time period T. Time taken to go from extreme to mid-amplitude position:Read the answer and worked solution →In SHM, when does kinetic energy equal potential energy?Read the answer and worked solution →A simple pendulum swings with period T on Earth. On Moon (g_moon = g/6), its period:Read the answer and worked solution →For SHM x = A sin(ωt), find ratio of kinetic energy at t = T/4 to t = T/8:Read the answer and worked solution →A particle's SHM is described by x = 3 sin(ωt) + 4 cos(ωt). Amplitude of resulting SHM:Read the answer and worked solution →Time period of a simple pendulum is T. If mass is doubled, period becomes:Read the answer and worked solution →A particle of mass 0.5 kg has SHM with amplitude 4 cm and time period 0.2 s. Find maximum kinetic energy:Read the answer and worked solution →A simple pendulum of length 1 m on Earth (g = 10) has frequency:Read the answer and worked solution →For SHM, the displacement-time graph is:Read the answer and worked solution →A mass m on a vertical spring of constant k. The equilibrium extension is x0 = mg/k. The mass is set into SHM. Frequency depends on:Read the answer and worked solution →Two simple pendulums of lengths 1 m and 4 m start swinging in phase. They will be in phase again after:Read the answer and worked solution →A particle has SHM with x = A cos(ωt). At t = 0, the particle is at:Read the answer and worked solution →Phase difference between displacement and velocity in SHM:Read the answer and worked solution →The total energy of SHM is proportional to:Read the answer and worked solution →Period and frequencyRead the answer and worked solution →Counting oscillationsRead the answer and worked solution →Period, frequency and unitsRead the answer and worked solution →The SHM displacementRead the answer and worked solution →Symbol to meaning in SHMRead the answer and worked solution →Angular frequencyRead the answer and worked solution →SHM and circular motionRead the answer and worked solution →Velocity and acceleration amplitudesRead the answer and worked solution →
Worked practice questions
A particle executing SHM has time period T. Time taken to go from extreme position (x = A) to x = A/2:
- T/4
- T/6
- T/8
- T/12
Show answer and solution
Correct answer: B. T/6
With x = A cos(ωt) starting at extreme: x = A/2 when cos(ωt) = 1/2, so ωt = π/3, hence t = T/(2π) × π/3 = T/6.
A simple pendulum of length 1 m has period T (g = 9.8). New length needed to double T:
- 0.5 m
- 1 m
- 4 m
- 2 m
Show answer and solution
Correct answer: C. 4 m
T ∝ √L. To double T, L must increase 4×.
So new L = 4 × 1 = 4 m.
(Original T ≈ 2.01 s; new T ≈ 4.02 s.)
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