The velocity of a certain simple harmonic oscillator is given by `v= -(12 m//s) sin [(6.0 "rad"//s)t]`. What is the amplitude of the simple harmonic motion?

Options

A. 2.0m
B. 4.0m
C. 6.0m
D. 8.0m

Answer

To find the amplitude of the simple harmonic motion (SHM) given the velocity equation \( v = -12 \sin(6.0 \, \text{rad/s} \cdot t) \), we can follow these steps:

### Step 1: Understand the relationship between velocity and displacement in SHM
In simple harmonic motion, the velocity \( v \) can be expressed in terms of displacement \( x \) and angular frequency \( \omega \). The general equations are:
- Displacement: \( x(t) = A \cos(\omega t + \phi) \)
- Velocity: \( v(t) = -A \omega \sin(\omega t + \phi) \)

Where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 2: Identify the parameters from the given velocity equation
From the given velocity equation:
\[ v = -12 \sin(6.0 \, \text{rad/s} \cdot t) \]

We can identify:
- \( A \omega = 12 \)
- \( \omega = 6.0 \, \text{rad/s} \)

### Step 3: Solve for the amplitude \( A \)
Using the relationship \( A \omega = 12 \), we can substitute the value of \( \omega \):
\[ A \cdot 6.0 = 12 \]

Now, solve for \( A \):
\[ A = \frac{12}{6.0} = 2 \, \text{m} \]

### Conclusion
The amplitude of the simple harmonic motion is \( 2 \, \text{m} \).

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