Question

Two equal masses are attached to separate identical springs next to one another. One mass is pulled so its spring stretches 20 cm and the other is pulled so its spring stretches only 10 cm. The masses are released simultaneously.Which mass reaches the equilibrium point first?

Answer 1

both reaches at the same time.

Step-by-step explanation:

The time period of the mass is given by

`T=2\pi \sqrt{(m)/(K)}`

where, K be the spring constant and m be the mass attached.

As both the masses are same and springs are identical so K is same for both the springs.

Thus, both the masses will reach at the equilibrium position at the same time.

Answer 2

Both reaches equilibrium simultaneously.

Step-by-step explanation:

Given that,

First stretch spring= 20 cm

Second stretch spring = 10 cm

Two equal masses are suspended from identical spring.

Their forces constants are equal time period for oscillation

Using formula of time period

`T=2\pi\sqrt{(m)/(k)}`

Where, T = time period

m = mass

k = spring constant

Time period independent of amplitude.

Hence, Both reaches equilibrium simultaneously.