A spring is placed in vertical position by suspending it from a hook at its top. A similar hook on the bottom of the spring is at 11 cm above a table top. A mass of 75 g and of negligible size is then suspended from the bottom hook, which is measured to be 4.5 cm above the table top. The mass is then pulled down a distance of 4 cm and released. Find the approximate. Position of the bottom hook after 5 s? Take `g=10(m)/(s^2)` and hook's mass to be negligible.

Options

A. 5 cm above the table top
B. 4.5 cm above the table top
C. 9 cm above the table top
D. 0.5 cm above the table top

Explanation

To solve the problem step by step, we will follow the outlined process below:

### Step 1: Understand the Setup
- The spring is suspended vertically with a hook at the top.
- The bottom hook is positioned 11 cm above the tabletop.
- A 75 g mass is suspended from the bottom hook, which is measured to be 4.5 cm above the tabletop.

### Step 2: Calculate the Initial Extension of the Spring
- The initial position of the mass is at 4.5 cm above the tabletop.
- The distance from the top hook to the bottom hook is 11 cm.
- Therefore, the initial extension \( x_0 \) of the spring when the mass is at equilibrium is:
\[ x_0 = 11 \text{ cm} - 4.5 \text{ cm} = 6.5 \text{ cm} = 0.065 \text{ m} \]

### Step 3: Calculate the Weight of the Mass
- The mass \( m \) is given as 75 g, which is \( 0.075 \) kg.
- The weight \( W \) of the mass is calculated using:
\[ W = mg = 0.075 \text{ kg} \times 10 \text{ m/s}^2 = 0.75 \text{ N} \]

### Step 4: Calculate the Spring Constant \( K \)
- At equilibrium, the weight of the mass is balanced by the spring force:
\[ W = Kx_0 \]
Rearranging gives:
\[ K = \frac{W}{x_0} = \frac{0.75 \text{ N}}{0.065 \text{ m}} \approx 11.54 \text{ N/m} \]

### Step 5: Calculate the Time Period of Oscillation
- The time period \( T \) of the oscillation is given by:
\[ T = 2\pi \sqrt{\frac{m}{K}} \]
Substituting the values:
\[ T = 2\pi \sqrt{\frac{0.075 \text{ kg}}{11.54 \text{ N/m}}} \approx 0.5 \text{ s} \]

### Step 6: Determine the Number of Complete Cycles in 5 Seconds
- The total time given is 5 seconds.
- The number of complete cycles \( n \) in 5 seconds is:
\[ n = \frac{5 \text{ s}}{T} = \frac{5 \text{ s}}{0.5 \text{ s}} = 10 \]

### Step 7: Determine the Position After 5 Seconds
- Since the mass starts at its maximum displacement (pulled down 4 cm from the equilibrium position) and completes 10 full cycles, it returns to the equilibrium position.
- The equilibrium position of the mass is at 4.5 cm above the tabletop, meaning the bottom hook is at:
\[ 11 \text{ cm} - 4.5 \text{ cm} = 6.5 \text{ cm} \text{ above the tabletop} \]

### Step 8: Final Position of the Bottom Hook
- After 5 seconds, the bottom hook will be at 6.5 cm above the tabletop.

### Conclusion
The approximate position of the bottom hook after 5 seconds is **6.5 cm above the tabletop**.