A mass m = 50 g is dropped on a vertical spring of spring constant 500 N/m from a height h= 10 cm as shown in figure (18-E14). The mass sticks to the spring and executes simple harmonic oscillations after that. A concave mirror of focal length 12 cm facing the mass is fixed with its principal axis coinciding with the line of motion of the mass, its pole being at a distance of 30 cm from the free end of the spring. Find the length in which the image of the mass oscillates.

Answer

To solve the problem, we will follow these steps:

### Step 1: Convert the mass and height to SI units
Given:
- Mass \( m = 50 \, \text{g} = 50 \times 10^{-3} \, \text{kg} = 0.05 \, \text{kg} \)
- Height \( h = 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} = 0.1 \, \text{m} \)

### Step 2: Calculate the compression of the spring when the mass sticks to it
Using the equilibrium condition when the mass sticks to the spring:
\[ mg = kx \]
Where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( k = 500 \, \text{N/m} \) (spring constant)
- \( x \) is the compression of the spring.

Substituting the values:
\[ 0.05 \times 10 = 500 \times x \]
\[ 0.5 = 500x \]
\[ x = \frac{0.5}{500} = 0.001 \, \text{m} = 0.1 \, \text{cm} \]

### Step 3: Determine the mean position of the mass
The mean position of the mass after it sticks to the spring is:
\[ \text{Mean Position} = \text{Distance from the pole} + \text{Compression} \]
Given that the distance from the pole of the mirror to the free end of the spring is \( 30 \, \text{cm} \):
\[ \text{Mean Position} = 30 \, \text{cm} + 0.1 \, \text{cm} = 30.1 \, \text{cm} \]

### Step 4: Apply the work-energy theorem to find maximum compression
Using the work-energy theorem:
\[ mg(h + \delta) - \frac{1}{2}k\delta^2 = 0 \]
Where \( \delta \) is the maximum compression. Rearranging gives:
\[ mg(h + \delta) = \frac{1}{2}k\delta^2 \]

Substituting the known values:
\[ 0.05 \times 10 \times (0.1 + \delta) = \frac{1}{2} \times 500 \times \delta^2 \]
\[ 0.5(0.1 + \delta) = 250\delta^2 \]
\[ 0.05 + 0.5\delta = 250\delta^2 \]
Rearranging gives:
\[ 250\delta^2 - 0.5\delta - 0.05 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( \delta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Where \( a = 250, b = -0.5, c = -0.05 \):
\[ \delta = \frac{-(-0.5) \pm \sqrt{(-0.5)^2 - 4 \times 250 \times (-0.05)}}{2 \times 250} \]
\[ = \frac{0.5 \pm \sqrt{0.25 + 50}}{500} \]
\[ = \frac{0.5 \pm \sqrt{50.25}}{500} \]
Calculating \( \sqrt{50.25} \approx 7.07 \):
\[ \delta = \frac{0.5 \pm 7.07}{500} \]
Taking the positive root:
\[ \delta \approx \frac{7.57}{500} \approx 0.01514 \, \text{m} = 1.514 \, \text{cm} \]

### Step 6: Calculate the extremities of oscillation
The maximum compression gives the first extreme position:
\[ \text{First Extreme Position} = 30.1 \, \text{cm} + 1.514 \, \text{cm} = 31.614 \, \text{cm} \]
The amplitude is:
\[ \text{Amplitude} = 1.514 \, \text{cm} \]
The second extreme position (upward):
\[ \text{Second Extreme Position} = 30.1 \, \text{cm} - 1.514 \, \text{cm} = 28.586 \, \text{cm} \]

### Step 7: Use the mirror formula to find the image positions
Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Where \( f = -12 \, \text{cm} \) (focal length of the concave mirror), and \( u \) is the object distance (negative for real objects).

For the first extreme position \( u = -31.614 \, \text{cm} \):
\[ \frac{1}{v} = \frac{1}{-12} - \frac{1}{-31.614} \]
Calculating gives:
\[ \frac{1}{v} = -\frac{1}{12} + \frac{1}{31.614} \approx -0.08333 + 0.03164 \approx -0.05169 \]
Thus,
\[ v \approx -19.38 \, \text{cm} \]

For the second extreme position \( u = -28.586 \, \text{cm} \):
\[ \frac{1}{v} = \frac{1}{-12} - \frac{1}{-28.586} \]
Calculating gives:
\[ \frac{1}{v} = -0.08333 + 0.03496 \approx -0.04837 \]
Thus,
\[ v \approx -20.62 \, \text{cm} \]

### Step 8: Calculate the range of image oscillation
The range of image oscillation is:
\[ \text{Range} = |v_1 - v_2| = |(-19.38) - (-20.62)| = |20.62 - 19.38| = 1.24 \, \text{cm} \]

### Final Answer
The length in which the image of the mass oscillates is **1.24 cm**.

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