Question

Consider the situation shown in figure (12−E10). Show that if the blocks are displaced slightly in opposite direction and released, they will execute simple harmonic motion. Calculate the time period.
Figure

Answer 1

Explanation:

• During simple harmonic motion, the system’s centre of mass must not change.
• So, on the left-hand side, if the block m by distance x moves towards right, the block on the right-hand side must also move towards left by distance x.  
• The spring has the total compression 2x.

If v is the block’s velocity, then we can write using energy method:

$12 k^{2} x^{2}+12 m v^{2}+12 m v^{2}=C$

On both the sides, by taking the derivative with respect to t, we obtain:

$2 m v d v d t+2 k \times 2 x d x d t=0$

⇒Time period, $T=2 \pi m 2 k.$

Answer 2

Time period T = 2π√m/2k

Explanation:

The center of mass of the system should not change during the motion. So, if the block ‘m’ on the left moves towards right by a distance ‘x’, the block on the right moves towards left by the same distance ‘x’. So, total compression of the spring is 2x.

By energy method, 1/2k(2x)² + 1/2 mv² + 1/2 mv² = C

2kx² + mv² = C

Taking derivative on both sides with respect to t, we get:

m * 2V dv/dt + 2K * 2x dx/dt = 0

ma + 2Kx = 0 (as v = dx/dt and a = dv/dt)

a/x = -2k/m

ω²  = -2k/m

There ω = √2k/m

Time period T = 2π√m/2k

Please find attached picture for the diagram.