Question

A 121 g mass is hanging on a 15 cm long massless spring of spring constant 31.1 N/m. What would be the time period of the oscillation of the spring? Express your answer in seconds (s)

Answer 1

The time period of oscillation for a 121 g mass hanging from a 15 cm long massless spring with a spring constant of 31.1 N/m is approximately 0.39 seconds.

Step-by-step explanation:

The question asks about the time period of oscillation for a mass hanging from a spring. To calculate the time period of a mass-spring system undergoing simple harmonic motion, we use the formula:

T = 2π√(m/k)

where:

• T is the time period of oscillation,
• m is the mass attached to the spring (converted to kilograms), and
• k is the spring constant.

For the given mass of 121 g (which is 0.121 kg) and a spring constant of 31.1 N/m, the time period T can be calculated as follows:

T = 2π√(0.121 kg / 31.1 N/m)

T ≈ 2π√(0.00389 kg·m/N)

T ≈ 2π√(0.00389)

T ≈ 2π(0.0624)

T ≈ 0.39 seconds

Therefore, the time period of the spring's oscillation is approximately 0.39 seconds.