A particle is performing SHM according to the equation `x=(3cm)sin((2pi)/(18)+(pi)/(6))`, where t is in seconds. The distance travelled by the particle in 39 s is

A particle is performing SHM according to the equation x=(3cm)sin((2pit)/(18)+(pi)/(6)) , where t is in seconds. The distance travelled by the particle in 39 s is

A particle executes SHM according to equation x=10(cm)cos[2pit+(pi)/(2)] , where t is in seconds. The magnitude of the velocity of the particle at t=(1)/(6)s will be :-

A particle executes SHM according to equation x=10(cm)cos[2pit+(pi)/(2)] , where t is in seconds. The magnitude of the velocity of the particle at t=(1)/(6)s will be :-

A particle executes SHM according to equation x=10(cm)cos[2pit+(pi)/(2)] , where t is in seconds. The magnitude of the velocity of the particle at t=(1)/(6)s will be :-

A particles executes SHM according to the equation y = 2 sin 2pit , where y is displacement in m and t is time in s. The distance covered by the particle in 4 s of its motion is

The position of a particle is defined by vec r = 2sin((pi)/(4))t hati + 2cos((pi)/(4))t hatj+3t hatk , where 't' is in seconds. Then distance travelled by particle in 2 second motions is. [use pi^(2) = 10]

The position of a particle is defined by vec r = 2sin((pi)/(4))t hati + 2cos((pi)/(4))t hatj+3t hatk , where 't' is in seconds. Then distance travelled by particle in 2 second motions is. [use pi^(2) = 10]

A particle executes simple harmonic motion according to the displacement equation y = 10 cos(2 pi t + (pi)/(6))cm where t is in second The velocity of the particle at t = 1/6 second will be

x - t equation of a particle in SHM is x = (4 cm)cos ((pi)/(2)t) Here , t is in seconds. Find the distance travelled by the particle in first three seconds.

x - t equation of a particle in SHM is x = (4 cm)cos ((pi)/(2)t) Here , t is in seconds. Find the distance travelled by the particle in first three seconds.