The equation of displacement due to a sound wave is s=`s_0`sin^(2)((wt-kx)). if the bulk modulus of the medium is `B`, then the equation of pressure variation due to that sound is

Options

A. `B k s_(0) sin(2 omega t-2 k x)`
B. `-B k s_(0) sin(2 omega t-2 k x)`
C. `B k s_(0)cos^(2) ( omegat-kx)`
D. `-B k s_(0)cos^(2) ( omegat-kx)`

Explanation

To find the equation of pressure variation due to a sound wave given the displacement equation, we can follow these steps:

### Step 1: Understand the given displacement equation
The displacement due to a sound wave is given by:
\[ s = s_0 \sin^2(\omega t - kx) \]
where:
- \( s_0 \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number,
- \( t \) is time,
- \( x \) is the position.

### Step 2: Differentiate the displacement equation
To find the pressure variation, we need to calculate the spatial derivative of the displacement \( s \):
\[ \frac{\partial s}{\partial x} = \frac{\partial}{\partial x} (s_0 \sin^2(\omega t - kx)) \]
Using the chain rule, we get:
\[ \frac{\partial s}{\partial x} = s_0 \cdot 2 \sin(\omega t - kx) \cdot \cos(\omega t - kx) \cdot (-k) \]
This simplifies to:
\[ \frac{\partial s}{\partial x} = -2ks_0 \sin(\omega t - kx) \cos(\omega t - kx) \]

### Step 3: Use the relationship between pressure and displacement
The pressure variation \( P \) in a sound wave is related to the displacement by the equation:
\[ P = -B \frac{\partial s}{\partial x} \]
where \( B \) is the bulk modulus of the medium. Substituting the derivative we found:
\[ P = -B \left(-2ks_0 \sin(\omega t - kx) \cos(\omega t - kx)\right) \]
This simplifies to:
\[ P = 2Bks_0 \sin(\omega t - kx) \cos(\omega t - kx) \]

### Step 4: Use the double angle identity
Using the double angle identity for sine, \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \):
\[ P = Bks_0 \sin(2(\omega t - kx)) \]

### Final Result
Thus, the equation of pressure variation due to the sound wave is:
\[ P = Bks_0 \sin(2\omega t - 2kx) \]