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(c) T/4
(b) w√(a²2-y²)
(a) π/4ω
(c) 1s
(b) α
(c) 2π t + α
(b) 0.15 m/s
(c) 90°
A particle is executing SHM with a period of T seconds and amplitude a meter. The shortest time it takes to reach a point a 2 m from its mean position in seconds is
(c) T/4
(c) T/4
See lessThe velocity of particle in SHM at displacement y from mean position is
(b) w√(a²2-y²)
(b) w√(a²2-y²)
See lessThe instantaneous displacement of a simple pendulum oscillator is given by x = A cos (ωt+π/4) . If speed will be maximum at time
(a) π/4ω
(a) π/4ω
See lessA particle is oscillating according to the equation x = 7 cos (0.5π t), where t is in second. The point moves from the position of equilibrium to maximum displacement in time
(c) 1s
(c) 1s
See lessThe equation of simple harmonic motion y = a sin (2π t + α) then its phase at time t = 0s is
(b) α
(b) α
See lessThe equation of simple harmonic motion y = a sin (2π t + α) then its phase at time t is
(c) 2π t + α
(c) 2π t + α
See less. If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2s, its maximum velocity is
(b) 0.15 m/s
(b) 0.15 m/s
See lessTwo equations of two SHM y = a Sin (ωt–α) and y = a Cos (ωt–α). The phase difference between the two is
(c) 90°
(c) 90°
See less