Question

A 30 g mass attached to a spring oscillates according to the equation x(t) = 5.0 cm cos((3.3 rads/s)t - 1.2 rads) What is the magnitude of the maximum velocity A. 17 m/s B. 0.05 m/s O C.5 m/s D. 0.17 m/s

Answer 1

Question

Answer - Option D is correct , the magnitude of maximum velocity = 0.17 m/sec.

Given Equation of oscillating spring x(t) = 5.0 cm Cos (( 3.3 Rads/sec)t -1.2 Rads), this equation give the displacement of osscialating spring.

We know that velocity is defend as rate change of velocity with respect to time, now velocity

V(t) = dx(t)/dt , Now taking differentiation of above equation with respect to time,

V(t) = dx(t)/dt = - 5x 3.3 Sin((3.3Rads/sec-)t-1.2 Rads) , because dcos(at) /dt = -a sin(at)

V(t) = -16.5 Sin((3.3 Rads/sec)t- 1.2 Rads) , We know that Sin values lie between -1 to +1 So maximum velocity when Sin((3.3Rads/sec)t-1.2) = -1.

now Maximum velocity V(t)_max = 16.5 cm/sec , Now maximum velocity in  m/sec

V(t)_max =16.5 cm/sec =16.5 x10^-2 m/esc =0.165 m/esc = 0.17 m/sec taking only two digits

Answer , 'Magnitude of maximum velocity = 0.17 m/sec '