If the displacement (x) and velocity (v) of a particle executing SHM are related through the expression `3v^(2)=30-x^(2)`. If the time period of the particle is `T=pisqrt(n)`, then what is the value of n?

Answer

To solve the problem, we start with the given expression relating the displacement \( x \) and velocity \( v \) of a particle executing simple harmonic motion (SHM):

\[ 3v^2 = 30 - x^2 \]

### Step 1: Express \( v^2 \) in terms of \( x \)

From the equation, we can isolate \( v^2 \):

\[ v^2 = \frac{30 - x^2}{3} \]

### Step 2: Use the formula for velocity in SHM

In SHM, the velocity \( v \) can also be expressed in terms of angular frequency \( \omega \) and amplitude \( A \):

\[ v^2 = \omega^2 (A^2 - x^2) \]

### Step 3: Set the two expressions for \( v^2 \) equal to each other

Now we can set the two expressions for \( v^2 \) equal:

\[ \frac{30 - x^2}{3} = \omega^2 (A^2 - x^2) \]

### Step 4: Rearrange the equation

Multiplying both sides by 3 to eliminate the fraction:

\[ 30 - x^2 = 3\omega^2 (A^2 - x^2) \]

Expanding the right side:

\[ 30 - x^2 = 3\omega^2 A^2 - 3\omega^2 x^2 \]

### Step 5: Collect like terms

Rearranging gives us:

\[ 30 = 3\omega^2 A^2 + (3\omega^2 - 1)x^2 \]

### Step 6: Identify constants

For this equation to hold for all \( x \), the coefficient of \( x^2 \) must be zero (since the left side is a constant). Therefore, we set:

\[ 3\omega^2 - 1 = 0 \]

Solving for \( \omega^2 \):

\[ 3\omega^2 = 1 \implies \omega^2 = \frac{1}{3} \]

### Step 7: Find \( \omega \)

Taking the square root:

\[ \omega = \frac{1}{\sqrt{3}} \]

### Step 8: Relate \( \omega \) to the time period \( T \)

The time period \( T \) is related to \( \omega \) by the formula:

\[ T = \frac{2\pi}{\omega} \]

Substituting \( \omega \):

\[ T = \frac{2\pi}{\frac{1}{\sqrt{3}}} = 2\pi \sqrt{3} \]

### Step 9: Compare with the given time period

The problem states that \( T = \pi \sqrt{n} \). Setting the two expressions for \( T \) equal:

\[ 2\pi \sqrt{3} = \pi \sqrt{n} \]

### Step 10: Solve for \( n \)

Dividing both sides by \( \pi \):

\[ 2\sqrt{3} = \sqrt{n} \]

Squaring both sides:

\[ 4 \cdot 3 = n \implies n = 12 \]

Thus, the value of \( n \) is:

\[ \boxed{12} \]