A particle performing SHM is found at its equilibrium position at t = 1s and it is found to have a speed 0.25 = m//s at t=2s. If the period of oscillation is 6. Calculate amplitude of oscillation.

A particle performing SHM is found at its eqilbrium at t=1 sec .and it is found to have a speed of 0.25 m//s at t=2sec . If the period of oscillation is 6sec . Calculate amplitude of oscillation-

A particle performing S.H.M. about equilibrium position. Then the velocity of the particle is

A block of mass m is connected to a spring of spring constant k as shown in Fig. The block is found at its equilibrium position t=1s and it has a velocity of +0.25(m)/(s) at t=2s . The time period of oscillation is 6 s. Based on the given information answer the following question: Q. the amplitude of oscillation is

A block of mass m is connected to a spring of spring constant k as shown in Fig. The block is found at its equilibrium position t=1s and it has a velocity of +0.25(m)/(s) at t=2s . The time period of oscillation is 6 s. Based on the given information answer the following question: Q. the amplitude of oscillation is

A block of mass m is connected to a spring of spring constant k as shown in Fig. The block is found at its equilibrium position t=1s and it has a velocity of +0.25(m)/(s) at t=2s . The time period of oscillation is 6 s. Based on the given information answer the following question: Q. Determine the velocity of particle at t=5s .

A block of mass m is connected to a spring of spring constant k as shown in Fig. The block is found at its equilibrium position t=1s and it has a velocity of +0.25(m)/(s) at t=2s . The time period of oscillation is 6 s. Based on the given information answer the following question: Q. Determine the velocity of particle at t=5s .

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is pi m//s . Amplitude of oscillation is (cos 45^(@) = 1/(sqrt(2)))

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is pi m//s . Amplitude of oscillation is (cos 45^(@) = 1/(sqrt(2)))

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is pi m//s . Amplitude of oscillation is (cos 45^(@) = 1/(sqrt(2)))

When a particle executing SHM is at a distance of 0.03 m from the equilibrium position, its acceleration is 0.296m*s^(-2) . What is the time period of oscillation of the particle?