Question:
The equation of a transverse wave is \( y = y_0 \sin 2\pi\left(ft - \frac{x}{\lambda}\right) \). If the maximum particle velocity be four times that of wave velocity then
The equation of a transverse wave is \( y = y_0 \sin 2\pi\left(ft - \frac{x}{\lambda}\right) \). If the maximum particle velocity be four times that of wave velocity then
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Always distinguish between particle velocity (the physical velocity of the medium's elements vibrating about their equilibrium) and wave velocity (the speed at which the wave profile/energy propagates through space). They are related via the slope of the wave: \( v_p = -v_w \frac{\partial y}{\partial x} \).
Updated On: May 28, 2026
- \( \lambda = \frac{\pi y_0}{4} \)
- \( \lambda = \frac{\pi y_0}{2} \)
- \( \lambda = \pi y_0 \)
- \( \lambda = 2\pi y_0 \)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
The given equation represents a transverse wave. We need to find the relationship for wavelength \( \lambda \) when the maximum particle velocity of the medium is four times the propagation velocity of the wave.
Step 2: Key Formula or Approach:
1. Standard wave equation:
\[ y = y_0 \sin(\omega t - kx) = y_0 \sin \left(2\pi f t - \frac{2\pi}{\lambda} x\right) \]
where \( y_0 \) is the amplitude, \( \omega = 2\pi f \) is the angular frequency, and \( k = \frac{2\pi}{\lambda} \) is the wave number.
2. Maximum particle velocity (\( v_{p,\text{max}} \)):
\[ v_{p,\text{max}} = A \omega = y_0 (2\pi f) \]
3. Wave velocity (\( v_w \)):
\[ v_w = f \lambda \]
Step 3: Detailed Explanation:
According to the given condition:
\[ v_{p,\text{max}} = 4 v_w \]
Substitute the expressions for \( v_{p,\text{max}} \) and \( v_w \):
\[ y_0 (2\pi f) = 4 (f \lambda) \]
Since frequency \( f \) is non-zero, we can cancel it from both sides:
\[ 2\pi y_0 = 4 \lambda \]
Solve for the wavelength \( \lambda \):
\[ \lambda = \frac{2\pi y_0}{4} = \frac{\pi y_0}{2} \]
Step 4: Final Answer:
The wavelength is \( \lambda = \frac{\pi y_0}{2} \).
The given equation represents a transverse wave. We need to find the relationship for wavelength \( \lambda \) when the maximum particle velocity of the medium is four times the propagation velocity of the wave.
Step 2: Key Formula or Approach:
1. Standard wave equation:
\[ y = y_0 \sin(\omega t - kx) = y_0 \sin \left(2\pi f t - \frac{2\pi}{\lambda} x\right) \]
where \( y_0 \) is the amplitude, \( \omega = 2\pi f \) is the angular frequency, and \( k = \frac{2\pi}{\lambda} \) is the wave number.
2. Maximum particle velocity (\( v_{p,\text{max}} \)):
\[ v_{p,\text{max}} = A \omega = y_0 (2\pi f) \]
3. Wave velocity (\( v_w \)):
\[ v_w = f \lambda \]
Step 3: Detailed Explanation:
According to the given condition:
\[ v_{p,\text{max}} = 4 v_w \]
Substitute the expressions for \( v_{p,\text{max}} \) and \( v_w \):
\[ y_0 (2\pi f) = 4 (f \lambda) \]
Since frequency \( f \) is non-zero, we can cancel it from both sides:
\[ 2\pi y_0 = 4 \lambda \]
Solve for the wavelength \( \lambda \):
\[ \lambda = \frac{2\pi y_0}{4} = \frac{\pi y_0}{2} \]
Step 4: Final Answer:
The wavelength is \( \lambda = \frac{\pi y_0}{2} \).
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