A block attached to a spring with unknown spring constant oscillates with a period of 2.0 s. What is the period if d. The spring constant is doubled? Parts a to d are independent questions, each referring to the initial situation.

Verified step by step guidance

1. Identify the formula that relates the period of oscillation of a spring-mass system to the spring constant. The formula is T = 2\pi \sqrt{\frac{m}{k}}, where T is the period, m is the mass, and k is the spring constant.
2. Understand that if the spring constant is doubled, the new spring constant becomes k' = 2k.
3. Substitute the new spring constant into the period formula to see how it affects the period. The new period T' can be calculated using T' = 2\pi \sqrt{\frac{m}{2k}}.
4. Simplify the expression for the new period by factoring out the constants and variables. This can be done by recognizing that \sqrt{\frac{1}{2}} can be factored out of the square root, leading to T' = \frac{T}{\sqrt{2}}.
5. Conclude that the new period T' is T divided by the square root of 2, indicating that the period decreases when the spring constant is doubled.

Key Concepts

Simple Harmonic Motion (SHM)

Simple Harmonic Motion is a type of periodic motion where an object oscillates around an equilibrium position. In SHM, the restoring force is directly proportional to the displacement from the equilibrium position and acts in the opposite direction. This motion is characterized by a constant period, which depends on the mass of the object and the spring constant.

Period of Oscillation

The period of oscillation is the time taken for one complete cycle of motion in a periodic system. For a mass-spring system, the period (T) is given by the formula T = 2π√(m/k), where m is the mass attached to the spring and k is the spring constant. This relationship shows that the period is influenced by both the mass and the spring constant.

Effect of Spring Constant on Period

Doubling the spring constant (k) affects the period of oscillation in a mass-spring system. According to the formula T = 2π√(m/k), if k is increased, the period T decreases, indicating that the system oscillates faster. Specifically, if the spring constant is doubled, the new period becomes T' = 2π√(m/(2k)), which is T/√2, showing a reduction in the period.