4) After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings in 136 s. What is the value of g on this planet? (10.7 m/s?)
Simple harmonic motion
Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards the mean position. We see many examples of SHM around us, common ones are the motion of a pendulum, spring and vibration of strings in musical instruments, and so on.
Simple Pendulum
A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
Oscillation
In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion takes place. In mathematics, oscillations can also be described as vibrations. The most common examples of oscillation that is seen in daily lives include the alternating current (AC) or the motion of a moving pendulum.
![### Example Problem: Calculating the Gravitational Acceleration on an Unknown Planet
**Problem Statement:**
- **Context**: After landing on an unfamiliar planet, a space explorer constructs a simple pendulum.
- **Pendulum Length**: 50.0 cm
- **Experiment**: The pendulum makes 100 complete swings in 136 seconds.
- **Objective**: Determine the value of gravitational acceleration (g) on this planet.
- **Reference Answer**: 10.7 m/s²
**Details and Calculation Steps:**
1. **Convert the Pendulum Length**:
- Convert the length from centimeters to meters.
- \( l = 50.0 \, \text{cm} = 0.50 \, \text{m} \)
2. **Calculate the Period (T) of One Swing**:
- Given 100 complete swings in 136 seconds.
- The period \( T = \frac{\text{total time}}{\text{number of swings}} = \frac{136 \, \text{s}}{100} = 1.36 \, \text{s} \)
3. **Use the Formula for the Period of a Simple Pendulum**:
\[
T = 2 \pi \sqrt{\frac{l}{g}}
\]
- Rearrange the formula to solve for \( g \):
\[
g = \frac{4\pi^2 l}{T^2}
\]
4. **Substitute the Given Values**:
\[
g = \frac{4\pi^2 \times 0.50 \, \text{m}}{(1.36 \, \text{s})^2}
\]
5. **Perform the Calculations**:
- \( 4 \pi^2 \approx 39.478 \)
- \( (1.36)^2 = 1.8496 \)
- So,
\[
g = \frac{39.478 \times 0.50}{1.8496} \approx 10.7 \, \text{m/s}^2
\]
**Conclusion:**
- The value of the gravitational acceleration (g) on this unknown planet is approximately \( 10.7 \, \text{m/s}^2 \), confirming the reference answer provided](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45c3f407-ef34-4ad6-a068-fef4425d0691%2F8a06f26c-9d3b-45df-b983-f2b746fb7f73%2F8pytkx_processed.png&w=3840&q=75)
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