A light sprial spring supports `200g` weight at its lower end oscillates with a period of `1s`. The weight that must be removed from the lower end to reduce the period to `0.5s` is

Options

A. `100g`
B. `50g`
C. `150g`
D. `200g`

Explanation

To solve the problem, we will use the formula for the period of oscillation of a mass-spring system. The period \( T \) is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

where \( m \) is the mass attached to the spring and \( k \) is the spring constant.

### Step 1: Understand the initial conditions
We know that:
- Initial mass \( m_1 = 200 \, \text{g} = 0.2 \, \text{kg} \)
- Initial period \( T_1 = 1 \, \text{s} \)

### Step 2: Write the equation for the initial period
Using the formula for the period, we have:

\[ T_1 = 2\pi \sqrt{\frac{m_1}{k}} \]

### Step 3: Understand the final conditions
We want to find the mass \( m_2 \) that results in a new period \( T_2 = 0.5 \, \text{s} \).

### Step 4: Write the equation for the final period
Using the same formula for the new period:

\[ T_2 = 2\pi \sqrt{\frac{m_2}{k}} \]

### Step 5: Set up the ratio of periods
To relate the two periods, we can set up the ratio:

\[ \frac{T_1}{T_2} = \sqrt{\frac{m_1}{m_2}} \]

### Step 6: Substitute the known values
Substituting the known values into the ratio:

\[ \frac{1}{0.5} = \sqrt{\frac{0.2}{m_2}} \]

### Step 7: Simplify the equation
This simplifies to:

\[ 2 = \sqrt{\frac{0.2}{m_2}} \]

### Step 8: Square both sides to eliminate the square root
Squaring both sides gives:

\[ 4 = \frac{0.2}{m_2} \]

### Step 9: Solve for \( m_2 \)
Rearranging gives:

\[ m_2 = \frac{0.2}{4} = 0.05 \, \text{kg} = 50 \, \text{g} \]

### Step 10: Calculate the weight to be removed
The weight that must be removed from the original mass is:

\[ \text{Weight to remove} = m_1 - m_2 = 200 \, \text{g} - 50 \, \text{g} = 150 \, \text{g} \]

### Final Answer
The weight that must be removed to reduce the period to \( 0.5 \, \text{s} \) is \( 150 \, \text{g} \).

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