Question

You attach a 1.50 kg block to a horizontal spring that is fixed at one end. You pull the block until the spring is stretched by 0.500 m and release it from rest. Assume the block slides on a horizontal surface with negligible friction. The block reaches a speed of zero again 0.500 s after release (for the first time after release). What is the maximum speed of the block (in m/s)?

Answer 1

Given,

mass, m = 1.5 kg

Initial stretch, x = 0.5 m

Time taken to reach the speed of zero for the first time after release, t = 0.5 s

Since, the mass is kept on a friction less horizontal surface, hence it will exhibit simple harmonic motion.

Thus,

Amplitude, A = 0.5 m

Time period is 2 times the time taken to reach the speed of zero for the first time

hence,

Time period, T = 1 s

Since frequency, f = 1/T

therefore, angular frequency, = 2f = 2/ T

                                                    = (2 * 3.14 ) / 1 = 6.28

Now, equation of SHM is

x(t) = A cos( t + )

Now, for = 0

x(t) = A cos( t )

v = dx(t) / dt = d/dt ( A cos( t ) )

                     = -A sin( t )

v(t) = -A sin( t )

Hence,

Max velocity = |-A| = 0.5 * 6.28

                      = 3.14 m/s