A load of 10 kg is suspended from the free end of a string, causing the springs length to extend by (Let x m) from its normal position.

The spring completes 50 oscillations in 20 seconds, resulting in a time period of 0.4 seconds for each oscillation. As a result, the angular frequency is 2/0.4= 5 Hz.

Hello candidate,

The mass of 10 kg is suspended from the free end of a string, which produces an extension of (Let x m), in the springs length from the normal position.

The spring complete 50 oscillation within 20 seconds, so the time period for one oscillation is equal to 0.4 seconds. So, the angular frequency is equal to 2π/0.4= 5π Hz.

Time period= 2π×√m/k, where m = 10kg and Time= 0.4 second. So, the value of k= 2.469 KN/m.

Hope it helps!!