Question

A meter stick, suspended at one end by a 0.33 m long light string, is set into oscillation. The acceleration of gravity is 9.8 m/s2. Determine the period of oscillation. Answer in units of s. Answer in units of s. By what percentage does this differ from a 0.83 m long simple pendulum? Answer in units of percent. Answer in units of percent.

Answer 1

The period of oscillation for the meter stick suspended by a 0.33 m long light string is approximately 0.58 s. The period of oscillation for the meter stick differs from that of a 0.83 m long simple pendulum by approximately -30.1%.

Step-by-step explanation:

The period of oscillation for a meter stick suspended by a 0.33 m long light string can be determined using the formula T = 2π√(l/g), where T is the period, l is the length of the string, and g is the acceleration due to gravity. Plugging in the values, we get T = 2π√(0.33/9.8) s. By calculating this expression, we find that the period of oscillation is approximately 0.58 s.

To find the percentage by which this differs from a 0.83 m long simple pendulum, we can use the formula: ΔP = (P_new - P_old)/P_old x 100%, where ΔP is the percentage difference, P_new is the new period, and P_old is the old period. Plugging in the values, we get ΔP = (0.58 - 0.83)/0.83 x 100%, which gives us approximately -30.1%. Therefore, the period of oscillation for the meter stick differs from that of the simple pendulum by approximately -30.1%.