In (Q. 24.) the tension in string is T and the linear mass density of string is `mu`. The ratio of magnitude of maximum velocity of particle and the magnitude of maximum acceleration is

Options

A. `(1)/(2pi)sqrt((mul^(2))/(T))`
B. `2pisqrt((mul^(2))/(T))`
C. `(1)/(2pi)sqrt((T)/(mul^(2)))`
D. `(1)/(4pi)sqrt((mul^(2))/(T))`

Explanation

To solve the problem, we need to find the ratio of the maximum velocity of a particle on the string to the maximum acceleration of that particle. We will use the properties of standing waves and the relationships between tension, mass density, and wave frequency.

### Step-by-Step Solution:

1. **Understand the Wave Equation**:
The displacement of a particle in a standing wave can be expressed as:
\[ y(x, t) = 2A \sin(kx) \cos(\omega t) \]
where \( A \) is the amplitude, \( k \) is the wave number, and \( \omega \) is the angular frequency.

2. **Calculate Maximum Velocity**:
The velocity \( v \) of a particle is given by the time derivative of the displacement:
\[ v = \frac{dy}{dt} = -2A \omega \sin(kx) \sin(\omega t) \]
The maximum velocity \( v_{\text{max}} \) occurs when both \( \sin(kx) \) and \( \sin(\omega t) \) are at their maximum value of 1:
\[ v_{\text{max}} = 2A \omega \]

3. **Calculate Maximum Acceleration**:
The acceleration \( a \) is the time derivative of the velocity:
\[ a = \frac{dv}{dt} = -2A \omega^2 \sin(kx) \cos(\omega t) \]
The maximum acceleration \( a_{\text{max}} \) occurs when \( \sin(kx) \) and \( \cos(\omega t) \) are both at their maximum value of 1:
\[ a_{\text{max}} = 2A \omega^2 \]

4. **Find the Ratio of Maximum Velocity to Maximum Acceleration**:
Now, we can find the ratio:
\[ \text{Ratio} = \frac{v_{\text{max}}}{a_{\text{max}}} = \frac{2A \omega}{2A \omega^2} = \frac{1}{\omega} \]

5. **Express Angular Frequency in Terms of Tension and Linear Mass Density**:
The angular frequency \( \omega \) can be expressed in terms of the tension \( T \) and linear mass density \( \mu \):
\[ \omega = 2\pi f \]
where \( f \) is the frequency. The frequency for a string fixed at both ends is given by:
\[ f = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \]
For the second harmonic (\( n = 2 \)):
\[ f = \frac{2}{2L} \sqrt{\frac{T}{\mu}} = \frac{1}{L} \sqrt{\frac{T}{\mu}} \]
Thus:
\[ \omega = 2\pi \left(\frac{1}{L} \sqrt{\frac{T}{\mu}}\right) = \frac{2\pi}{L} \sqrt{\frac{T}{\mu}} \]

6. **Substituting Back into the Ratio**:
Now substituting \( \omega \) back into the ratio:
\[ \text{Ratio} = \frac{1}{\omega} = \frac{L}{2\pi} \sqrt{\frac{\mu}{T}} \]

### Final Answer:
The ratio of the magnitude of maximum velocity of the particle to the magnitude of maximum acceleration is:
\[ \frac{L}{2\pi} \sqrt{\frac{\mu}{T}} \]