Advertisements
Advertisements
प्रश्न
The displacement of a particle from its mean position (in metre) is given by, y = 0.2 sin(10 πt + 1.5π) cos(10 πt + 1.5π).
The motion of particle is ____________.
पर्याय
periodic but not S.H.M
non-periodic
simple harmonic motion with period 0.1 s
simple harmonic motion with period 0.2 s
MCQ
रिकाम्या जागा भरा
उत्तर
The displacement of a particle from its mean position (in metre) is given by, y = 0.2 sin(10 πt + 1.5π) cos(10 πt + 1.5π).
The motion of particle is simple harmonic motion with period 0.1 s.
Explanation:
`y = 0.2 sin(10 pit + 1.5 pi) cos( 10 pit + 1.5 pi)`
`= 0.1 sin2(10 pit + 1.5 pi) ......[ because sin2A = 2 sinA cosA]`
`= 0.1 sin(20 pit + 3.0 pi)`
`therefore "Time period", T = (2pi)/omega= (2pi)/(20pi) = 1/10 = 0.1 s`
shaalaa.com
Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?