A transverse sinusoidal wave is generted at one end of long, horizontal string by a bar that moves up and down through a distance of `1.00 cm`. The motion is continuous and is repreated regularly `120` times per second. The string has linear density `90 gm//m` and is kept under a tension of `900 N`. Find :
What is the transverse displacement `y`(in `cm`) when the minimum power transfer occurs
[Leave the answer in terms of `pi` wherever it occurs]

Options

A. `0`
B. `(1)/(2)`
C. `(1)/(4)`
D. `1`

Answer

To solve the problem, we need to find the transverse displacement \( y \) when the minimum power transfer occurs in a transverse sinusoidal wave on a string. Let's break down the solution step by step.

### Step 1: Understand the Given Information
- The amplitude of the wave generated is \( 1.00 \, \text{cm} \).
- The frequency of the wave is \( 120 \, \text{Hz} \).
- The linear density of the string is \( 90 \, \text{g/m} = 0.09 \, \text{kg/m} \).
- The tension in the string is \( 900 \, \text{N} \).

### Step 2: Calculate the Amplitude
The amplitude \( A \) of the wave is half the distance the bar moves up and down:
\[ A = \frac{1.00 \, \text{cm}}{2} = 0.50 \, \text{cm} = 0.005 \, \text{m} \]

### Step 3: Write the Wave Equation
The equation of a transverse wave can be expressed as:
\[ y(x, t) = A \sin(\omega t - kx) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number.

### Step 4: Calculate Angular Frequency \( \omega \)
The angular frequency \( \omega \) is given by:
\[ \omega = 2\pi f \]
Substituting \( f = 120 \, \text{Hz} \):
\[ \omega = 2\pi \times 120 = 240\pi \, \text{rad/s} \]

### Step 5: Calculate Wave Number \( k \)
The wave number \( k \) is given by:
\[ k = \frac{\omega}{v} \]
where \( v \) is the wave speed, calculated as:
\[ v = \sqrt{\frac{T}{\mu}} \]
Here, \( T = 900 \, \text{N} \) and \( \mu = 0.09 \, \text{kg/m} \):
\[ v = \sqrt{\frac{900}{0.09}} = \sqrt{10000} = 100 \, \text{m/s} \]
Now substituting \( v \):
\[ k = \frac{240\pi}{100} = 2.4\pi \, \text{rad/m} \]

### Step 6: Determine the Condition for Minimum Power Transfer
Power \( P \) is given by:
\[ P = T \frac{dy}{dt} \]
To find the minimum power transfer, we need \( \frac{dy}{dt} = 0 \). This occurs when:
\[ \cos(\omega t - kx) = 0 \]
This implies:
\[ \omega t - kx = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]
For the first instance, we take \( n = 0 \):
\[ \omega t - kx = \frac{\pi}{2} \]

### Step 7: Calculate the Transverse Displacement \( y \)
Substituting \( \omega t - kx = \frac{\pi}{2} \) into the wave equation:
\[ y = A \sin\left(\frac{\pi}{2}\right) = A \cdot 1 = A \]
Thus:
\[ y = 0.005 \, \text{m} = 0.50 \, \text{cm} \]

### Final Answer
The transverse displacement \( y \) when the minimum power transfer occurs is:
\[ \boxed{0.50 \, \text{cm}} \]