An object is attached to the bottom of a light verticle spring and set vibrating The maximum speed of the object is `15cm//sec` and the time in centimeters is

Options

A. `3.0`
B. `2.0`
C. `1.5`
D. `1.0`

Explanation

To solve the problem, we need to find the amplitude of the oscillation of the object attached to the spring, given the maximum speed (v_max) and the time period (T). The maximum speed in simple harmonic motion is related to the amplitude and the time period by the formula:

\[ v_{max} = A \cdot \omega \]

where \( \omega \) is the angular frequency given by:

\[ \omega = \frac{2\pi}{T} \]

Thus, we can rewrite the formula for maximum speed as:

\[ v_{max} = A \cdot \frac{2\pi}{T} \]

From this equation, we can solve for the amplitude \( A \):

\[ A = \frac{v_{max} \cdot T}{2\pi} \]

### Step-by-Step Solution:

1. **Identify the given values**:
- Maximum speed \( v_{max} = 15 \, \text{cm/s} \) (convert to meters: \( 15 \, \text{cm/s} = 0.15 \, \text{m/s} \))
- Time period \( T \) (not given, but we will express the amplitude in terms of \( T \))

2. **Substitute the values into the amplitude formula**:
\[ A = \frac{0.15 \cdot T}{2\pi} \]

3. **Calculate the amplitude**:
- Since we do not have a specific value for \( T \), we leave the amplitude in terms of \( T \):
\[ A = \frac{0.15 \cdot T}{2\pi} \]

4. **Final expression**:
- The amplitude \( A \) is expressed as:
\[ A = \frac{0.15T}{2\pi} \, \text{meters} \]