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The path difference between the two wave...

The path difference between the two waves
`y_(1)=a_(1) sin(omega t-(2pi x)/(lambda)) and y(2)=a_(2) cos(omega t-(2pi x)/(lambda)+phi)` is

A

`(lambda)/(2pi)/phi`

B

`(lambda)/(2pi)(phi+(pi)/(2))`

C

`(2pi)/(lambda)(phi-(pi)/(2))`

D

`(2pi)/(lambda)(phi)`

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The correct Answer is:
To find the path difference between the two waves given by the equations \( y_1 = a_1 \sin\left(\omega t - \frac{2\pi x}{\lambda}\right) \) and \( y_2 = a_2 \cos\left(\omega t - \frac{2\pi x}{\lambda} + \phi\right) \), we can follow these steps: ### Step 1: Rewrite the cosine function We can express the cosine function in terms of sine using the identity: \[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \] Thus, we can rewrite \( y_2 \) as: \[ y_2 = a_2 \sin\left(\omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2}\right) \] ### Step 2: Identify the phase of each wave Now we can identify the phase of each wave: - For \( y_1 \): The phase is \( \phi_1 = \omega t - \frac{2\pi x}{\lambda} \). - For \( y_2 \): The phase is \( \phi_2 = \omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2} \). ### Step 3: Calculate the phase difference The phase difference \( \Delta \phi \) between the two waves is given by: \[ \Delta \phi = \phi_2 - \phi_1 \] Substituting the values we found: \[ \Delta \phi = \left(\omega t - \frac{2\pi x}{\lambda} + \phi + \frac{\pi}{2}\right) - \left(\omega t - \frac{2\pi x}{\lambda}\right) \] This simplifies to: \[ \Delta \phi = \phi + \frac{\pi}{2} \] ### Step 4: Relate phase difference to path difference The path difference \( \Delta x \) can be related to the phase difference using the formula: \[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \] So, we can rearrange this to find \( \Delta x \): \[ \Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} \] Substituting \( \Delta \phi = \phi + \frac{\pi}{2} \): \[ \Delta x = \frac{\left(\phi + \frac{\pi}{2}\right) \cdot \lambda}{2\pi} \] ### Final Result Thus, the path difference between the two waves is: \[ \Delta x = \frac{\lambda}{2\pi} \left(\phi + \frac{\pi}{2}\right) \]

To find the path difference between the two waves given by the equations \( y_1 = a_1 \sin\left(\omega t - \frac{2\pi x}{\lambda}\right) \) and \( y_2 = a_2 \cos\left(\omega t - \frac{2\pi x}{\lambda} + \phi\right) \), we can follow these steps: ### Step 1: Rewrite the cosine function We can express the cosine function in terms of sine using the identity: \[ \cos(\theta) = \sin\left(\theta + \frac{\pi}{2}\right) \] Thus, we can rewrite \( y_2 \) as: ...
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