A ball on a massless, rigid rod oscillates as a simple pendulum with a period of 3.0 s on Earth. If the ball is replaced with another ball having twice the mass and bring to the moon that has g value 1/6 of Earth, the period will be around O 7.3 s 18 s 5.2 s 6.0 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.
![**Problem Description:**
A ball on a massless, rigid rod oscillates as a simple pendulum with a period of 3.0 seconds on Earth. If the ball is replaced with another ball having twice the mass and brought to the Moon, where the gravitational acceleration (g) is 1/6 that of Earth, determine the new period of oscillation.
**Multiple Choice Options:**
- 7.3 s
- 18 s
- 5.2 s
- 6.0 s
**Analysis:**
For a simple pendulum, the period \( T \) is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. Note that the period does not depend on the mass of the ball. On the Moon, the gravity \( g_{\text{moon}} = \frac{1}{6}g_{\text{earth}} \).
Substituting the gravity on the Moon:
\[ T_{\text{moon}} = 2\pi \sqrt{\frac{L}{g_{\text{moon}}}} = 2\pi \sqrt{\frac{L}{\frac{1}{6}g_{\text{earth}}}} = 2\pi \sqrt{6 \cdot \frac{L}{g_{\text{earth}}}} \]
\[ T_{\text{moon}} = \sqrt{6} \cdot T_{\text{earth}} \]
With \( T_{\text{earth}} = 3.0 \) s:
\[ T_{\text{moon}} = \sqrt{6} \cdot 3.0 \approx 7.3\, \text{s} \]
Thus, the correct answer is 7.3 s.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7d688b34-c43a-4a2a-9e75-79997b42342d%2F06e969b0-ae0c-4b00-b09a-2dda289df83d%2F59wpybr_processed.jpeg&w=3840&q=75)
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