A particle in SHM has a maximum speed of 30 cm/s and a maximum acceleration of 90 cm/s². The period of oscillation is:
A$\pi/3$ s
B$2\pi/3$ s
C$\pi$ s
D$2\pi$ s
Answer & Solution
Correct answer: B. $2\pi/3$ s
$v_{max} = A\omega = 30$; $a_{max} = A\omega^2 = 90$. Dividing: $\omega = a_{max}/v_{max} = 90/30 = 3$ rad/s. So $T = 2\pi/\omega = 2\pi/3$ s.
Related questions
A particle of mass 0.2 kg performs S.H.M. of amplitude 0.05 m with period π s. Its maximumIn damped harmonic motion, the amplitude of oscillationTwo springs of force constants k₁ and k₂ are connected in series and stretched by an exterA particle in S.H.M. of period 2 s starts from rest at the extreme position. Its displacemThe kinetic energy of a particle of mass m in linear S.H.M. of amplitude A at displacementA simple pendulum of length 1 m on the earth's surface (g = π² m/s²) has periodA spring-mass system of mass m and spring constant k oscillates with periodA particle executes S.H.M. with amplitude A and angular frequency ω. Its maximum accelerat