A block of mass m = 2.00 kg resting on a horizontal frictionless surface is atta

Question

A block of mass m = 2.00 kg resting on a horizontal frictionless surface is attached to a horizontal spring of negligible mass and spring constant k = 450 N/m. The other end of the spring is attached to a wall that does not move. At time t = 0, the spring is stretched so that the block has displacement x_o = 6.00 cm from the equilibrium point and the system released. a) Calculate the period of the resulting oscillations and determine how many oscillations the system makes over a time period Delta t = 13.0 sec b) Find the maximum speed that the block reaches as it oscillates back and forth. c) Determine the positive value of x, the displacement of the block from the equilibrium position, when the elastic potential energy of the block and its kinetic energy are equal. d) Find the earliest time at which the elastic potential energy and kinetic energy of the block are equal and calculate the block's speed at this time.

Explanation / Answer

a)

period of oscillations :

T = 2*pi*sqrt(m/k)

= 2*pi*sqrt(2/450)

= 0.419 s

Number of oscillations = 13/0.419 = 31

b)

Maximum speed = sqrt(k*x^2/m)

= sqrt(450*0.06^2/2)

= 0.9 m/s

c)

When PE = KE

0.5*k*x^2 = 0.5*mv^2

So, 450*x^2 = 2*v^2 ------ (1)

Now, the change in KE = 0.5*2*v^2 = 0.5*(450)*(0.06^2 - x^2) -------- (2)

So, v = 0.636 m/s , x = 0.042 m <-------------answer

d)

x = A*cos(2*pi*t/T)

So, for x = 0.042 = 0.06*cos(2*pi*t/0.419)

So, t = 3.03 s <-------answer

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