Exam Review Guidance: Physics Concepts and Problem-Solving

Q4. Simple Harmonic Motion – A 5 kg mass on a spring exhibits simple harmonic motion as shown in the position vs. time plot. Each grid line is 1 s in the time axis and 1 cm in the position axis.

Background

Topic: Simple Harmonic Motion (SHM)

This question tests your understanding of SHM, including force, spring constant, and energy calculations using position vs. time graphs.

Key Terms and Formulas:

• Amplitude ($A$): Maximum displacement from equilibrium.

• Spring constant ($k$): Measures stiffness of the spring.

• Hooke's Law: $F = -kx$

• Period ($T$): Time for one complete oscillation.

• Angular frequency ($\omega$): $\omega = \frac{2\pi}{T}$

• SHM equation: $x(t) = A \cos(\omega t + \phi)$

• Spring-mass system: $\omega = \sqrt{\frac{k}{m}}$

• Kinetic energy at position $x$: $K = \frac{1}{2}k(A^2 - x^2)$

Step-by-Step Guidance

1. Examine the position vs. time plot to determine the amplitude ($A$) and period ($T$) of the oscillation. Count the number of grid squares for maximum displacement and for one full cycle.

   

2. Use the period ($T$) to find the angular frequency ($\omega$) with $\omega = \frac{2\pi}{T}$.

3. Apply the relationship $\omega = \sqrt{\frac{k}{m}}$ to solve for the spring constant $k$. Rearranged: $k = m\omega^2$.

4. For kinetic energy at $x = \frac{A}{4}$, use $K = \frac{1}{2}k(A^2 - x^2)$. Substitute $x = \frac{A}{4}$ and your value for $k$.

Try solving on your own before revealing the answer!

Final Answer:

Spring constant $k \approx 3.08\,\text{N/m}$

Kinetic energy at $x = \frac{A}{4}$: $K \approx 1.45 \times 10^{-4}\,\text{J}$

We used the period from the graph to find $\omega$, then calculated $k$ using the mass. For kinetic energy, we substituted $x = \frac{A}{4}$ into the energy formula.