Question

Consider a block-spring system as shown in the figure. The block is attached to the spring, which is attached to the inside wall of the box. The mass of the block is 0.2kg. Suppose you pull the block 10cm to the right and release it. The angular frequency for it to oscillate is 30 rad/s. Neglect friction between the block and the bottom of the box. (a)What is the spring constant of the spring? (b)What will be the amplitude of the oscillations? (c)Taking to the right to be positive, at what point in the oscillation is the velocity minimum and what is its minimum value? (d)At what point in the oscillation is the acceleration minimum, and what is its minimum value? (e)What is the total energy of the spring-block system? (f)If you take t = 0 to be the instant when you release the block, write an equation of motion for the oscillation, x(t) =?, identifying the values of all constants that you use. (g)Imagine now that the box, with the spring and block in it, starts moving to the left with an acceleration a = 4 m/s 2 . By how much does the equilibrium position of the block shift (relative to the box), and in what direction?

Answer 1

a] Angular frequency and spring's constant are related through:

so,

=> k = 180 N/m

b] The block is pulled 10 cm to the right initially.

So, the amplitude of oscillations is: A = 10 cm = 0.1m

c] The velocity is minimum at x = A = 0.1m and its value there is v = 0 m/s

at this point, all the kinetic energy of the mass is converted to elastic potential energy.

d] The acceleration is minimum when the velocity is maximum and this happens when the block is at x = 0 m (center). Here, the acceleration is a = 0 m/s². At this point, all energy is in the form of kinetic energy.