A mass-spring system oscillates such that the mass moves on a rough surface having coefficient of friction `mu`. It is compressed by a distance a from its normal length and, on being released, it moves to a distance b from its equilibrium position. The decrease in amplitude for one half-cycle (-a to b) is

Options

A. `(mu mg)/(k)`
B. `(2mu mg)/(k)`
C. `(mu g)/(k)`
D. `(k)/(mumg)`

Explanation

To solve the problem of the decrease in amplitude for a mass-spring system oscillating on a rough surface, we can follow these steps:

### Step 1: Understand the System
We have a mass-spring system where the mass is placed on a rough surface with a coefficient of friction \( \mu \). The spring is compressed by a distance \( a \) from its normal length and then released. The mass moves to a distance \( b \) from its equilibrium position.

### Step 2: Identify the Energy in the System
The potential energy stored in the spring when compressed by distance \( a \) is given by:
\[ E_a = \frac{1}{2} k a^2 \]
where \( k \) is the spring constant.

When the mass moves to distance \( b \), the potential energy in the spring is:
\[ E_b = \frac{1}{2} k b^2 \]

### Step 3: Calculate Work Done Against Friction
The work done against friction as the mass moves from position \( a \) to position \( b \) can be calculated using the formula:
\[ W = f \cdot d \]
where \( f \) is the frictional force and \( d \) is the distance moved. The frictional force \( f \) is given by:
\[ f = \mu m g \]
The distance \( d \) that the mass moves is \( a + b \):
\[ W = \mu m g (a + b) \]

### Step 4: Apply Conservation of Energy
According to the conservation of energy, the initial potential energy stored in the spring will be equal to the work done against friction plus the potential energy at position \( b \):
\[ \frac{1}{2} k a^2 = \frac{1}{2} k b^2 + \mu m g (a + b) \]

### Step 5: Rearranging the Equation
Rearranging the above equation gives:
\[ \frac{1}{2} k a^2 - \frac{1}{2} k b^2 = \mu m g (a + b) \]
Factoring out the left side:
\[ \frac{1}{2} k (a^2 - b^2) = \mu m g (a + b) \]

### Step 6: Factor the Left Side
Using the difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]
Thus, we can rewrite the equation as:
\[ \frac{1}{2} k (a - b)(a + b) = \mu m g (a + b) \]

### Step 7: Cancel \( (a + b) \)
Assuming \( a + b \neq 0 \), we can cancel \( (a + b) \) from both sides:
\[ \frac{1}{2} k (a - b) = \mu m g \]

### Step 8: Solve for Decrease in Amplitude
Now, solving for \( (a - b) \):
\[ a - b = \frac{2 \mu m g}{k} \]
This \( (a - b) \) represents the decrease in amplitude for one half-cycle.

### Final Result
The decrease in amplitude for one half-cycle is:
\[ \Delta A = a - b = \frac{2 \mu m g}{k} \]