The maximum velocity of a particle, executing S.H.M. with an amplitude 7 mm is 4.4 m s⁻¹ The period of oscillation is [π = 22/7] (a) 100 s (b) 10 s (c) 0.1 s (d) 0.01 s

Answer: (d). vmax = Aω = (2π A)/(T) ⇒ T = 2π Avmax T = 2×22/7× 7×10⁻³4.4 = (0.044)/(4.4) T = 0.01 s So the answer is (d) 0.01 s.

The maximum velocity of a particle, executing S.H.M. with an… | MHT CET 2024 (15 May, Shift 1)

Q23MHT CET 2024 · 15 MAY · SHIFT 1Easy

Velocity & Acceleration in SHM

The maximum velocity of a particle, executing S.H.M. with an amplitude 7 mm is $4.4\text{ m s}^{-1}$ The period of oscillation is $\left[\pi = \frac{22}{7}\right]$

Show answer and solution

Answer: (d)

$v_{max} = A\omega = \frac{2\pi A}{T} \Rightarrow T = \frac{2\pi A}{v_{max}}$
$T = \frac{2\times\frac{22}{7}\times 7\times10^{-3}}{4.4} = \frac{0.044}{4.4}$
$T = 0.01$ s
So the answer is (d) 0.01 s.

Report a mistake

More questions on Velocity & Acceleration in SHM

A particle performing S.H.M. with maximum velocity ' V '. If the amplitude double and…MHT CET 2024 (10 May, Shift 1)
The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is…MHT CET 2024 (4 May, Shift 2)
A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s…JEE Main 2024 (6 Apr, Shift 1)
A particle performing S.H.M. starts from equilibrium position and its time period is 12…MHT CET 2024 (11 May, Shift 1)
The velocity of particle executing S.H.M. varies with displacement (x) as 4V² = 50 - x²…MHT CET 2024 (4 May, Shift 1)
The displacement of a particle executing SHM is given by x = 10sin (ω t + π/3) m. The…JEE Main 2024 (4 Apr, Shift 2)
A particle performs simple harmonic motion with amplitude A. Its speed is increased to…JEE Main 2024 (31 Jan, Shift 1)
A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean…JEE Main 2024 (27 Jan, Shift 1)

Velocity & Acceleration in SHM question from MHT CET 2024 (15 May, Shift 1)

More questions on Velocity & Acceleration in SHM