A pendulum has a period of 1.95 s on Earth. Part A What is its period on Mars, where the acceleration of gravity is about 0.37 that on Earth? Express your answer to two significant figures and include the appropriate units. µA TMars = Value Units
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.
![### Pendulum Period Calculation
**Given:**
A pendulum has a period of 1.95 seconds on Earth.
**Part A:**
**Problem:**
What is its period on Mars, where the acceleration of gravity is about 0.37 times that on Earth?
**Task:**
Express your answer to two significant figures and include the appropriate units.
**Input Field for Solution:**
- \( T_{\text{Mars}} = \) [Value] [Units]
### Explanation:
To calculate the period of a pendulum on Mars, we use the formula for the period of a simple pendulum:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
- \( T \) is the period
- \( L \) is the length of the pendulum
- \( g \) is the acceleration due to gravity
Since the length \( L \) is constant, and we know the period on Earth \( T_{\text{Earth}} \) and \( g_{\text{Earth}} \), we can set up the proportion:
\[ \frac{T_{\text{Mars}}}{T_{\text{Earth}}} = \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Mars}}}} \]
Given:
- \( T_{\text{Earth}} = 1.95 \) seconds
- \( g_{\text{Mars}} = 0.37 \times g_{\text{Earth}} \)
By substituting the values into the equation, solve for \( T_{\text{Mars}} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F54a9d33b-c075-4af5-8013-3f49004b10ba%2Fbc6d23cd-a606-453c-8d23-ef3f43b622bd%2Flzz50z_processed.png&w=3840&q=75)
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