CW Spring Torque Spring Table B Part 3: Hanging Force of Moment G81 Force Elongation Torque Constant gravity mass Arm 260a 2621.0490cm 832.9N39.7cm
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.
How to get the Spring constant and the force of gravity?
![**Transcription for Educational Website**
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**Table A Part 3: CCW Summation of Torque**
| | Mass | \( \times 9.7 \) Force | Moment arm | Torque |
|----------------|------------|----------------|------------|----------|
| Suspended Mass | 174.53 g | 1712.14 N | 95 cm | 1626.5 J |
| Meter Stick | 195.51 g | 1917.95 N | 20 cm | 383.65 J |
| Spring Clamp | 27.5 g | 220.72 N | 5 cm | 11.045 J |
**Summation of Torques = 2021.045 J**
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**Table B Part 3: CW Spring Torque**
| Torque | Moment Arm | Force | Elongation | Spring Constant | Hanging Mass | Force of Gravity | % Diff |
|--------|------------|---------|------------|-----------------|---------------|------------------|--------|
| 2021.04 | 90 cm | 882.9 N | 39.7 cm | | 260 g | | |
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**Calculations:**
\[ 152.03 + 22.5 = 174.53 g \]
\[ \begin{align*}
F & = 195.51 \times 9.81 \, \text{m/s}^2 = 1917.95 \, \text{N} \\
& = 22.5 \times 9.81 \, \text{m/s}^2 = 220.72 \, \text{N} \\
& = 1712.14 \, \text{N} \times 0.95 = 1626.5 \, \text{T} \\
& = 1917.95 \, \text{N} \times 0.20 = 383.65 \, \text{T} \\
& = 220.72 \, \text{N} \times 0.05 = 11.045 \, \text{T}
\end{align*} \]
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**Questions:**
1) **In Part 1, why was the mass of the meter stick NOT involved in the determination of torques?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8631c00b-fe9e-4ed1-8720-e6ab9879b041%2F437716d1-612e-43d8-895a-a53108de80b6%2Fwhwtan.jpeg&w=3840&q=75)
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