A particle having mass 10 g oscillates according to the equation `x=(2.0cm) sin[(100s^-1)t+pi/6]`. Find a the amplitude the time period and the spring constant b. the position, the velocity and the acceleration at t=0.

Explanation

To solve the problem step by step, we will break it down into two parts:

### Part A: Finding Amplitude, Time Period, and Spring Constant

1. **Identify the Amplitude (A)**:
- The given equation of motion is \( x = (2.0 \, \text{cm}) \sin[(100 \, \text{s}^{-1})t + \frac{\pi}{6}] \).
- The amplitude \( A \) is the coefficient of the sine function.
- Therefore, \( A = 2.0 \, \text{cm} \).

2. **Calculate the Angular Frequency (\(\omega\))**:
- From the equation, we see that \( \omega = 100 \, \text{s}^{-1} \).

3. **Find the Time Period (T)**:
- The time period \( T \) is related to angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]
- Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{100} = \frac{\pi}{50} \, \text{s} \]

4. **Calculate the Spring Constant (K)**:
- The relationship between angular frequency, mass, and spring constant is given by:
\[ \omega = \sqrt{\frac{K}{m}} \]
- Rearranging gives:
\[ K = \omega^2 \cdot m \]
- Convert mass from grams to kilograms:
\[ m = 10 \, \text{g} = 0.01 \, \text{kg} \]
- Now substituting the values:
\[ K = (100)^2 \cdot 0.01 = 10000 \cdot 0.01 = 100 \, \text{N/m} \]

### Summary of Part A:
- Amplitude \( A = 2.0 \, \text{cm} \)
- Time Period \( T = \frac{\pi}{50} \, \text{s} \)
- Spring Constant \( K = 100 \, \text{N/m} \)

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### Part B: Finding Position, Velocity, and Acceleration at \( t = 0 \)

1. **Find the Position (x) at \( t = 0 \)**:
- Substitute \( t = 0 \) into the displacement equation:
\[ x = 2 \sin(100 \cdot 0 + \frac{\pi}{6}) = 2 \sin(\frac{\pi}{6}) = 2 \cdot \frac{1}{2} = 1 \, \text{cm} \]

2. **Calculate the Velocity (v) at \( t = 0 \)**:
- The velocity is the derivative of the position:
\[ v = \frac{dx}{dt} = 2 \cdot 100 \cdot \cos(100t + \frac{\pi}{6}) \]
- At \( t = 0 \):
\[ v = 200 \cos(\frac{\pi}{6}) = 200 \cdot \frac{\sqrt{3}}{2} = 100\sqrt{3} \, \text{cm/s} \]

3. **Calculate the Acceleration (a) at \( t = 0 \)**:
- The acceleration is the derivative of the velocity:
\[ a = \frac{dv}{dt} = -2 \cdot 100^2 \sin(100t + \frac{\pi}{6}) \]
- At \( t = 0 \):
\[ a = -20000 \sin(\frac{\pi}{6}) = -20000 \cdot \frac{1}{2} = -10000 \, \text{cm/s}^2 \]

### Summary of Part B:
- Position at \( t = 0 \): \( x = 1 \, \text{cm} \)
- Velocity at \( t = 0 \): \( v = 100\sqrt{3} \, \text{cm/s} \)
- Acceleration at \( t = 0 \): \( a = -10000 \, \text{cm/s}^2 \)

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