The period of the pendulum T (in s) is given below, where L-length of pendulum (in m), g-9.8 m/s². - 2 √ T= 2TA 1. Draw a free body diagrams of a pendulum while in motion. Rf = restoring force T= tension w = weight T 0. 2000 吧 W Data and Calculations Table 1 Mass (kg) 0.0500 0.1000 T₁(s) 1.9175 S 1.9293 1.9261s T₂(s) 1.91445 1.9911 s 1.9376 s 7 Error Length L= 87 cm = 0.87 m T3(s) Taverage(s) 1.9139 1.9153 s 1.92785 1.94945 1.9305 1.9314. 1.9153-2 x 100 2 = -4.24 S -4.24 -2.53 -3.43 @1.9494-2x100 2 = -2.53 S 2. Calculate the period T for the length that was used Tstandard- punto à from the stope %error=ITaverage -Tstandard I/Tstandard x 100=_ % Error 5. Does the period depend on the mass of the pendulum? Explain.

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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Hello, can you please help me do part 2
**Understanding Pendulum Motion: Calculations and Analysis**

The period of a pendulum \( T \) (in seconds) is determined by the formula:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

where:
- \( L \) = Length of the pendulum (in meters)
- \( g \) = Acceleration due to gravity (9.8 m/s\(^2\))

1. **Conceptual Diagram**

   - A free body diagram of a pendulum in motion is illustrated. 
   - Forces acting on the pendulum:
     - \( T \) = Tension in the string
     - \( W \) = Weight of the pendulum
     - \( RF \) = Restoring force

2. **Data and Calculations - Table 1**

   | Mass (kg) | \( T_1 \)(s) | \( T_2 \)(s) | \( T_3 \)(s) | \( T_{\text{average}} \) (s) | % Error |
   |-----------|-------------|-------------|-------------|------------------------------|---------|
   | 0.0500    | 1.9175      | 1.9144      | 1.9139      | 1.9153                       | -4.24   |
   | 0.1000    | 1.9293      | 1.9911      | 1.9278      | 1.9494                       | -2.53   |
   | 0.2000    | 1.9261      | 1.9376      | 1.9305      | 1.9314                       | -3.43   |

3. **Experimental Setup**

   - Length \( L \) = 87 cm = 0.87 m

4. **Calculations**

   - **Period \( T \)**: Calculate \( T \) for the length used. 
   
   - **Percentage Error**: \(\% \text{error} = \frac{|T_{\text{average}} - T_{\text{standard}}|}{T_{\text{standard}}} \times 100\)

5. **Analysis**

   - Does the period depend on the mass of the pendulum?

     - **Conclusion**: The period is a function of \( L \) and \( g \
Transcribed Image Text:**Understanding Pendulum Motion: Calculations and Analysis** The period of a pendulum \( T \) (in seconds) is determined by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where: - \( L \) = Length of the pendulum (in meters) - \( g \) = Acceleration due to gravity (9.8 m/s\(^2\)) 1. **Conceptual Diagram** - A free body diagram of a pendulum in motion is illustrated. - Forces acting on the pendulum: - \( T \) = Tension in the string - \( W \) = Weight of the pendulum - \( RF \) = Restoring force 2. **Data and Calculations - Table 1** | Mass (kg) | \( T_1 \)(s) | \( T_2 \)(s) | \( T_3 \)(s) | \( T_{\text{average}} \) (s) | % Error | |-----------|-------------|-------------|-------------|------------------------------|---------| | 0.0500 | 1.9175 | 1.9144 | 1.9139 | 1.9153 | -4.24 | | 0.1000 | 1.9293 | 1.9911 | 1.9278 | 1.9494 | -2.53 | | 0.2000 | 1.9261 | 1.9376 | 1.9305 | 1.9314 | -3.43 | 3. **Experimental Setup** - Length \( L \) = 87 cm = 0.87 m 4. **Calculations** - **Period \( T \)**: Calculate \( T \) for the length used. - **Percentage Error**: \(\% \text{error} = \frac{|T_{\text{average}} - T_{\text{standard}}|}{T_{\text{standard}}} \times 100\) 5. **Analysis** - Does the period depend on the mass of the pendulum? - **Conclusion**: The period is a function of \( L \) and \( g \
Expert Solution
Step 1: Determine the given data:

straight L equals 87 space cm equals 0.87 space straight m
Average space values colon
straight T subscript 1 equals 1.9153 space sec
straight T subscript 2 equals 1.9494 space sec
straight T subscript 3 equals 1.9314 space sec

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