An object of mass m, oscillating on the end of a spring with spring constant k has amplitude A. its maximum speed is

Options

A. `A sqrt(k//m)`
B. `A^(2)k//m`
C. `A sqrt(m//k)`
D. `A m//k`

Answer

To find the maximum speed of an object of mass \( m \) oscillating on the end of a spring with spring constant \( k \) and amplitude \( A \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**: The object is undergoing simple harmonic motion (SHM). The displacement \( x \) of the object can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Differentiate to Find Velocity**: The velocity \( v(t) \) of the object is the derivative of the displacement with respect to time:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]

3. **Determine Maximum Speed**: The maximum speed occurs when \( \cos(\omega t + \phi) \) is at its maximum value, which is 1. Therefore, the maximum speed \( v_{\text{max}} \) can be expressed as:
\[ v_{\text{max}} = A \omega \]

4. **Relate Angular Frequency to Spring Constant**: The angular frequency \( \omega \) for a mass-spring system is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
where \( k \) is the spring constant and \( m \) is the mass of the object.

5. **Substitute for Maximum Speed**: Now, substitute \( \omega \) into the equation for maximum speed:
\[ v_{\text{max}} = A \sqrt{\frac{k}{m}} \]

6. **Final Expression**: Thus, the maximum speed of the object is:
\[ v_{\text{max}} = A \sqrt{\frac{k}{m}} \]

### Conclusion:
The maximum speed of the object oscillating on the end of the spring is given by:
\[ v_{\text{max}} = A \sqrt{\frac{k}{m}} \]