Question

The end point of a spring vibrates with a period of 2.5 seconds when a mass m is attached to it. when this mass is increased by 9.294×10^-1 kg, the period is found to be 3.8 seconds. Find the value of m.

a) 0.5 kg
b) 1.2 kg
c) 2.0 kg
d) 3.5 kg

Answer 1

To find the value of mass 'm' that alters the period of a spring-mass system from 2.5 to 3.8 seconds when increased by 0.9294 kg, we apply the formula for the period of simple harmonic motion and solve the resulting equation to get a mass of approximately 1.2 kg.

Step-by-step explanation:

The question you've asked pertains to simple harmonic motion (SHM) and we can use the formula for the period T of a mass-spring system, which is T = 2π√(m/k), where m is the mass attached to the spring and k is the spring constant. To solve for the mass m, we need to set up two equations based on the two given periods and solve them simultaneously. The first equation is based on the initial period of 2.5 seconds with mass m, and the second is based on the period of 3.8 seconds with mass m + 0.9294 kg.

Leveraging these two equations:

• T1 = 2π√(m/k) with T1 = 2.5 s
• T2 = 2π√((m + 0.9294 kg)/k) with T2 = 3.8 s

We divide the second equation by the first to eliminate k and solve for m:

(3.8/2.5)^2 = (m + 0.9294 kg)/m

Solving the equation yields m approximately equal to 1.2 kg, hence option b) is correct.