A block on a frictionless surface is attached to a spring and undergoes simple h
Practice Questions
Q1
A block on a frictionless surface is attached to a spring and undergoes simple harmonic motion. If the spring constant is 200 N/m and the mass is 2 kg, what is the period of oscillation?
0.5 s
1 s
2 s
4 s
Questions & Step-by-Step Solutions
A block on a frictionless surface is attached to a spring and undergoes simple harmonic motion. If the spring constant is 200 N/m and the mass is 2 kg, what is the period of oscillation?
Step 1: Identify the formula for the period of oscillation T in simple harmonic motion, which is T = 2π√(m/k).
Step 2: Identify the values given in the problem: the mass m = 2 kg and the spring constant k = 200 N/m.
Step 3: Substitute the values into the formula: T = 2π√(2/200).
Step 4: Simplify the fraction inside the square root: 2/200 = 0.01.
Step 5: Take the square root of 0.01: √(0.01) = 0.1.
Step 6: Multiply by 2π: T = 2π(0.1).
Step 7: Calculate the final result: T = 0.2π, which is approximately 0.63 seconds.
Simple Harmonic Motion – The motion of an object attached to a spring, characterized by a restoring force proportional to the displacement from equilibrium.
Period of Oscillation – The time taken for one complete cycle of motion, calculated using the formula T = 2π√(m/k).
Spring Constant – A measure of the stiffness of the spring, denoted by k, which affects the period of oscillation.
Mass – The quantity of matter in the block, denoted by m, which influences the period of oscillation.
