Physics 2 - Form 2

Q1. Define the following terms as applied to simple harmonic motion:
i. amplitude
ii. period

B. What is a simple pendulum? Give two examples of systems that can be used to perform simple harmonic motion.

C. A body of mass 0.30 kg is made to perform simple harmonic motion with a speed of 2.50 m/s and an amplitude of 0.04 m. Calculate:
i. the maximum displacement of the particle,
ii. the acceleration of the particle,
iii. the energy associated with the motion.

D. A small bob of mass [tex]\(20 \times 10^{-2} \text{ kg}\)[/tex] oscillates as a simple pendulum with an amplitude of [tex]\(5.0 \times 10^{-2} \text{ m}\)[/tex] and a period of 2 s. Calculate the maximum velocity of the bob.



Answer :

Sure, I'll provide a detailed, step-by-step solution to solve the given problems related to simple harmonic motion.

### Part A: Definition of Terms

1. Amplitude:
- Amplitude is the maximum displacement of a particle from its equilibrium position in simple harmonic motion (SHM). It represents the peak value of the oscillation.

2. Period:
- The period of a simple harmonic oscillator is the time taken for the particle to complete one full cycle of the motion. It is typically denoted by [tex]\( T \)[/tex] and measured in seconds.

### Part B: Simple Pendulum and Examples

1. Simple Pendulum:
- A simple pendulum consists of a mass (called the bob) suspended from a fixed point by a string or a rod of length [tex]\( l \)[/tex], which swings back and forth under the influence of gravity. The motion is periodic and can be considered as simple harmonic for small angles of displacement.

2. Examples of Systems Performing Simple Harmonic Motion:
- A mass attached to a horizontal spring oscillating back and forth.
- The swinging of a simple pendulum (for small angles).

### Part C: Calculations for a Body Performing SHM

Given:
- Mass, [tex]\( m = 0.30 \)[/tex] kg
- Speed, [tex]\( v = 2.50 \)[/tex] m/s
- Amplitude, [tex]\( A = 0.04 \)[/tex] m

1. Maximum Kinetic Energy:
- The kinetic energy in SHM is maximum when the particle is at the equilibrium position (where all the energy is kinetic).
- [tex]\( KE_{\text{max}} = \frac{1}{2} m v^2 \)[/tex]
- Substituting the given values:
[tex]\[ KE_{\text{max}} = \frac{1}{2} \times 0.30 \, \text{kg} \times (2.50 \, \text{m/s})^2 = 0.9375 \, \text{J} \][/tex]

2. Maximum Acceleration:
- Maximum acceleration occurs at the maximum displacement (amplitude) in SHM. It can be calculated using:
- [tex]\( a_{\text{max}} = \frac{v^2}{A} \)[/tex]
- Substituting the given values:
[tex]\[ a_{\text{max}} = \frac{(2.50 \, \text{m/s})^2}{0.04 \, \text{m}} = 156.25 \, \text{m/s}^2 \][/tex]

3. Total Energy Associated with the Motion:
- In SHM, the total mechanical energy (E) is the sum of kinetic and potential energies, and it remains constant throughout the motion.
- The total energy is equal to the maximum kinetic energy:
[tex]\[ E_{\text{total}} = KE_{\text{max}} = 0.9375 \, \text{J} \][/tex]

### Part D: Maximum Velocity of a Bob of a Pendulum

Given:
- Mass of the bob, [tex]\( m = 20 \times 10^{-2} \)[/tex] kg = 0.20 kg
- Amplitude, [tex]\( A = 5.0 \times 10^{-2} \)[/tex] m = 0.05 m
- Period, [tex]\( T = 2.1 \)[/tex] s

1. Maximum Velocity:
- The maximum velocity in simple harmonic motion can be found using the relationship:
- [tex]\( v_{\text{max}} = A \omega \)[/tex]
- Here, [tex]\( \omega \)[/tex] is the angular frequency, given by:
- [tex]\( \omega = \frac{2\pi}{T} \)[/tex]
- Substituting [tex]\( T \)[/tex]:
[tex]\[ \omega = \frac{2\pi}{2.1 \, \text{s}} \approx 2.992 \, \text{rad/s} \][/tex]
[tex]\[ v_{\text{max}} = 0.05 \, \text{m} \times 2.992 \, \text{rad/s} \approx 0.1496 \, \text{m/s} \][/tex]

In summary:
- The maximum kinetic energy is [tex]\( 0.9375 \)[/tex] J.
- The maximum acceleration is [tex]\( 156.25 \, \text{m/s}^2 \)[/tex].
- The total energy associated with the motion is [tex]\( 0.9375 \)[/tex] J.
- The maximum velocity of the bob is [tex]\( 0.1496 \, \text{m/s} \)[/tex].