In a transverse wave the distance between a crest and neighbouring trough at the same instant is 4.0 cm and the distance between a crest and trought at the same place is 1.0cm. The next crest appears at the same place after a time interval of 0.4s. The maximum speed of the vibrating particles in the medium is:

Options

A. `(3 pi)/2 cm//s`
B. `(5 pi)/2 cm//s`
C. `pi/2 cm//s`
D. `2 pi cm//s`

Answer

To find the maximum speed of the vibrating particles in the medium for the given transverse wave, we can follow these steps:

### Step 1: Identify the Given Information
- Distance between a crest and neighboring trough = 4.0 cm
- Distance between a crest and trough at the same place = 1.0 cm
- Time interval for the next crest to appear = 0.4 s

### Step 2: Determine the Amplitude
The distance between a crest and trough is half the wavelength. Therefore, the amplitude \( A \) can be calculated as:
- Distance between crest and trough = \( \frac{1}{2} \) wavelength
- Thus, the amplitude \( A = \frac{1.0 \, \text{cm}}{2} = 0.5 \, \text{cm} \)

### Step 3: Calculate the Angular Frequency
The time interval for the next crest to appear is the time period \( T \):
- \( T = 0.4 \, \text{s} \)

The angular frequency \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{0.4} = 5\pi \, \text{rad/s} \]

### Step 4: Calculate the Maximum Speed
The maximum speed \( v_{\text{max}} \) of the vibrating particles in a transverse wave can be calculated using the formula:
\[ v_{\text{max}} = A \cdot \omega \]
Substituting the values of \( A \) and \( \omega \):
\[ v_{\text{max}} = 0.5 \, \text{cm} \cdot 5\pi \, \text{rad/s} \]
Calculating this gives:
\[ v_{\text{max}} = \frac{5\pi}{2} \, \text{cm/s} \]

### Final Answer
The maximum speed of the vibrating particles in the medium is:
\[ \frac{5\pi}{2} \, \text{cm/s} \]

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