Question

Consider a block-spring system inside a box, 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 0.2 kg. For parts (a) through (f), assume that the box does not move. Suppose you pull the block 10 cm to the right and release it. The angular frequency of the oscillations is 30 rad/s. Neglect friction between the block and the bottom of the box.

(a) What is the spring constant?

(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/s2 . By how much does the equilibrium position of the block shift (relative to the box), and in what direction?

Answer 1

It is a case of the horizontal spring-mass system performing SHM.

GIVEN: A block of mass, m = 0.2 Kg is attached to the spring as shown in the figure. T whole system is placed in a fixed box with the inner walls of which the spring is attached.

The spring stretched 10 cm to the right as shown and released.and thus performs oscillations.

Friction is zero between the block and the bottom of the box, = 0

The angular frequency of the oscillations, = 30 rad/sec.

SOLUTION:

Here mean position of the block is at the natural length of the spring

(a) In the above case, we know that when the spring is stretched, a restoring spring force acts on it and is given by,

Here, K - spring constant,

x - displacement/ length to which the spring is stretched.

On comparing the above equation with F = m a, we get,

= F

  Therefore, the acceleration, 'a' is given by

  

On comparing the above acceleration equation with that of SHM, we get

  .............(here, the negative sign means the acceleration is always toward the mean position)

  

  

   (The spring constant)

(b) Since in the above case, there's no non- conservative force is acting, thus, we can apply the law of conservation of mechanical energy, i.e.,

...................(a)

Kinitial and Kfinal are the initial and final kinetic energies respectively

Uinitial  and Ufinal are the initial and final potential energies respectively.

At time, t = 0, x = +10 cm, when mass was released, i.e., initial velocity, v0 = 0

At maximum displacement, i.e., amplitude (A), the velocity is zero i.e., vfinal = 0

Therefore, the equation (a) becomes,

  

   

(c)

In SHM, velocity is given by,

For velocity to be minimum, displacement, 'x' from the mean position must be maximum, i.e., equal to amplitude, i.e.,  x = A

Therefore,

  

i.e., minimum velocity is zero at amplitude, i.e., at maximum distance from the mean position.

(d)

We know that the acceleration of the above oscillating system is given by,

  

For the minimum value of the acceleration, the value of displacement from the mean position of the block must be minimum,  

i.e., xmin = 0

Thus, the minimum acceleration will be amin = 0 at the minimum displacement x =0 from the mean position.