One end of a string of length L is tied to the ceiling of a lift accelerating upwards with an acceleration 2g. The other end o the string is free. The linear mass density of the string varies linearly from 0 to `lamda` from bottom to top. The acceleration of a wave pulled through out the string is `(pg)/(4)`. Find p.

Answer

To solve the problem step by step, we will analyze the situation involving the string in the lift and derive the necessary equations to find the value of \( p \).

### Step 1: Understand the Setup
We have a string of length \( L \) tied to the ceiling of a lift that is accelerating upwards with an acceleration of \( 2g \). The linear mass density of the string varies linearly from \( 0 \) at the bottom to \( \lambda \) at the top.

### Step 2: Define the Linear Mass Density
The linear mass density \( \mu(y) \) at a distance \( y \) from the bottom of the string can be expressed as:
\[ \mu(y) = \frac{\lambda}{L} y \]
where \( y \) varies from \( 0 \) (bottom) to \( L \) (top).

### Step 3: Calculate the Mass of a Small Element
Consider a small element of the string of length \( dy \) at a height \( y \). The mass \( dm \) of this element is given by:
\[ dm = \mu(y) \cdot dy = \left(\frac{\lambda}{L} y\right) dy \]

### Step 4: Calculate the Total Mass of the Element
To find the total mass \( m(y) \) of the string from the bottom to height \( y \), we integrate:
\[ m(y) = \int_0^y \mu(y') dy' = \int_0^y \left(\frac{\lambda}{L} y'\right) dy' = \frac{\lambda}{L} \int_0^y y' dy' = \frac{\lambda}{L} \cdot \frac{y^2}{2} = \frac{\lambda y^2}{2L} \]

### Step 5: Analyze Forces on the Element
The tension \( T \) in the string at height \( y \) must balance the weight of the mass below it and provide the necessary force for the upward acceleration of \( 2g \):
\[ T - m(y)g = m(y) \cdot (2g) \]
Substituting \( m(y) \):
\[ T - \left(\frac{\lambda y^2}{2L}\right) g = \left(\frac{\lambda y^2}{2L}\right) \cdot (2g) \]
This simplifies to:
\[ T = 3 \left(\frac{\lambda y^2}{2L}\right) g \]

### Step 6: Calculate Wave Velocity
The velocity \( v \) of a wave on the string is given by:
\[ v = \sqrt{\frac{T}{\mu(y)}} \]
Substituting \( T \) and \( \mu(y) \):
\[ v = \sqrt{\frac{3 \left(\frac{\lambda y^2}{2L}\right) g}{\frac{\lambda}{L} y}} = \sqrt{\frac{3g y}{2}} \]

### Step 7: Relate Wave Velocity to Acceleration
From the problem, we know that the acceleration of the wave is given as \( \frac{pg}{4} \). The wave acceleration can also be expressed in terms of the wave velocity:
\[ a = \frac{d(v^2)}{dy} = \frac{d}{dy}\left(\frac{3gy}{2}\right) = \frac{3g}{2} \]

### Step 8: Equate the Two Expressions for Acceleration
Setting the two expressions for acceleration equal gives:
\[ \frac{3g}{2} = \frac{pg}{4} \]
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ \frac{3}{2} = \frac{p}{4} \]

### Step 9: Solve for \( p \)
Multiplying both sides by \( 4 \):
\[ p = 6 \]

### Final Answer
Thus, the value of \( p \) is:
\[ \boxed{6} \]