A mass m attached to a spring of spring constant `k` is stretched a distance `x_0` from its equilibrium position and released with no initial velocity. The maximum speed attained by mass in its subsequent motion and the time at which this speed would be attained are, respectively.

Options

A. (a)`sqrt((k)/(m))x_(0)` and `sqrt((m)/(k))`
B. (b)`sqrt((k)/(m))(x_(0))/(2)` and `(pi)/(2)sqrt((m)/(k))`
C. (c)`sqrt((k)/(m))x_(0)` and `(pi)/(2)sqrt((m)/(k))`
D. (d)`sqrt((k)/(m))(x_(0))/(2)` and `pisqrt((m)/(k))`

Explanation

To solve the problem, we need to find the maximum speed attained by a mass \( m \) attached to a spring with spring constant \( k \) when it is stretched a distance \( x_0 \) from its equilibrium position and released with no initial velocity. We also need to determine the time at which this maximum speed is attained.

### Step-by-Step Solution:

**Step 1: Determine the potential energy at the maximum displacement.**
- When the mass is stretched to \( x_0 \), the potential energy stored in the spring is given by:
\[ PE = \frac{1}{2} k x_0^2 \]

**Hint for Step 1:** Remember that potential energy in a spring is calculated using the formula \( \frac{1}{2} k x^2 \).

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**Step 2: Relate potential energy to kinetic energy at the mean position.**
- As the mass moves towards the mean position, all the potential energy converts into kinetic energy at the mean position (where the speed is maximum):
\[ KE = \frac{1}{2} m v_{max}^2 \]
- Setting the potential energy equal to the kinetic energy:
\[ \frac{1}{2} k x_0^2 = \frac{1}{2} m v_{max}^2 \]

**Hint for Step 2:** Use the principle of conservation of energy, where potential energy converts to kinetic energy.

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**Step 3: Solve for maximum speed \( v_{max} \).**
- Canceling \( \frac{1}{2} \) from both sides and rearranging gives:
\[ k x_0^2 = m v_{max}^2 \]
\[ v_{max}^2 = \frac{k}{m} x_0^2 \]
- Taking the square root:
\[ v_{max} = x_0 \sqrt{\frac{k}{m}} \]

**Hint for Step 3:** When solving for \( v_{max} \), remember to take the square root of both sides.

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**Step 4: Determine the time at which maximum speed occurs.**
- The time period \( T \) of the motion is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- The maximum speed occurs at a quarter of the time period, which is:
\[ t = \frac{T}{4} = \frac{1}{4} \times 2\pi \sqrt{\frac{m}{k}} = \frac{\pi}{2} \sqrt{\frac{m}{k}} \]

**Hint for Step 4:** Remember that the maximum speed occurs at \( \frac{1}{4} \) of the time period.

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### Final Answers:
- The maximum speed attained by the mass is:
\[ v_{max} = x_0 \sqrt{\frac{k}{m}} \]
- The time at which this speed is attained is:
\[ t = \frac{\pi}{2} \sqrt{\frac{m}{k}} \]