Question

A 0.50 kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. When the spring is 4.0 cm shorter than its equilibrium length, the speed of the block is 0.50 m/s, The greatest speed of the block is:

Answer 1

This is SHM.

k = 80 N/m            m = 0.50 kg
ω = angular frequency of SHM of spring mass system
=> ω²  =  k/m = 160         => ω = 4√10 rad/sec²

Let the equation of SHM be:    x = x₀  Sin (4√10 t)
                                    and    v = x₀ * 4√10  Cos( 4√10 t)

Given  that  displacement  0.04 meters = x₀ Sin(4√10  t)
and  speed:                        0.50 m/s = x₀ 4√10 * Cos (4√10 t)

Hence,   0.04² + (0.50/4√10)² = x₀²
                 x₀ = 0.0562  meters   or 5.62 cm

Thus the maximum speed of the block = x₀ * ω 
      = 0.711 m/sec

Answer 2

This is SHM.

k = 80 N/m            m = 0.50 kg
ω = angular frequency of SHM of spring mass system
=> ω²  =  k/m = 160         => ω = 4√10 rad/sec²

Let the equation of SHM be:    x = x₀  Sin (4√10 t)
                                    and    v = x₀ * 4√10  Cos( 4√10 t)

Given  that  displacement  0.04 meters = x₀ Sin(4√10  t)
and  speed:                        0.50 m/s = x₀ 4√10 * Cos (4√10 t)

Hence,   0.04² + (0.50/4√10)² = x₀²
                 x₀ = 0.0562  meters   or 5.62 cm

Thus the maximum speed of the block = x₀ * ω 
      = 0.711 m/sec