Question

A rectangular block of mass 5 kg attached to an horizontal spiral spring executes simple harmonic motion of amplitude 1 m and time period 3.14s. The maximum force exerted by spring on block is _____N.

Answer 1

Correct Answer: 20

To solve this problem, we need to determine the maximum force exerted by the spring on the block when the system is executing simple harmonic motion (SHM). The force exerted by a spring in SHM is given by Hooke's Law: F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position. The maximum force occurs when the displacement is at its maximum value, which is the amplitude of the motion.

First, we determine the spring constant, k. The time period, T, of the SHM is related to k and the mass, m, by the formula T = 2π√(m/k). Rearranging for k, we get k = (4π²m)/T².

Substituting the given values: m = 5 kg, T = 3.14 s, we find:

k = (4π² × 5)/(3.14)² ≈ (4 × 9.8696 × 5)/9.8596 ≈ 20 N/m

Now, using the spring constant k and the amplitude A = 1 m, we calculate the maximum force:

F_max = kA = 20 × 1 = 20 N

Therefore, the maximum force exerted by the spring on the block is 20 N. This value is within the expected range of 20,20.