If the equation of transverse wave is `y=5 sin 2 pi [(t)/(0.04)-(x)/(40)]`, where distance is in cm and time in second, then the wavelength of the wave is

Options

A. 60 cm
B. 40 cm
C. 35 cm
D. 25 cm

Answer

To find the wavelength of the wave given by the equation \( y = 5 \sin \left( 2\pi \left( \frac{t}{0.04} - \frac{x}{40} \right) \right) \), we can follow these steps:

### Step 1: Identify the wave equation format
The general form of a wave equation is:
\[ y = A \sin(\omega t - kx) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number.

### Step 2: Rewrite the given equation
The given equation can be rewritten to identify \( \omega \) and \( k \):
\[ y = 5 \sin \left( 2\pi \left( \frac{t}{0.04} - \frac{x}{40} \right) \right) \]
This can be expanded as:
\[ y = 5 \sin \left( \frac{2\pi t}{0.04} - \frac{2\pi x}{40} \right) \]

### Step 3: Identify \( \omega \) and \( k \)
From the rewritten equation, we can identify:
- \( \omega = \frac{2\pi}{0.04} \)
- \( k = \frac{2\pi}{40} \)

### Step 4: Calculate the wave number \( k \)
Now, we can calculate \( k \):
\[ k = \frac{2\pi}{40} = \frac{\pi}{20} \, \text{cm}^{-1} \]

### Step 5: Relate wave number to wavelength
The wave number \( k \) is related to the wavelength \( \lambda \) by the formula:
\[ k = \frac{2\pi}{\lambda} \]
Setting the two expressions for \( k \) equal gives:
\[ \frac{\pi}{20} = \frac{2\pi}{\lambda} \]

### Step 6: Solve for wavelength \( \lambda \)
To find \( \lambda \), we can cross-multiply:
\[ \pi \lambda = 40\pi \]
Dividing both sides by \( \pi \):
\[ \lambda = 40 \, \text{cm} \]

### Conclusion
The wavelength of the wave is \( \lambda = 40 \, \text{cm} \).

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