The period of oscillation of a suspended thin cylindrical magnet is 4 seconds. It is broken into exactly two halves. Find the period of oscillation of each half when freely suspended.

Options

A. 4s
B. 2s
C. 1s
D. `2sqrt2s`

Answer

To find the period of oscillation of each half of a cylindrical magnet after it has been broken into two equal halves, we can follow these steps:

### Step 1: Understand the formula for the period of oscillation
The period of oscillation \( T \) of a magnet can be expressed using the formula:
\[ T = 2\pi \sqrt{\frac{I}{MB}} \]
where:
- \( I \) is the moment of inertia of the magnet,
- \( M \) is the magnetic moment,
- \( B \) is the magnetic field.

### Step 2: Identify the initial conditions
We know that the original period of oscillation of the thin cylindrical magnet is \( T = 4 \) seconds.

### Step 3: Break the magnet into two halves
When the magnet is broken into two equal halves:
- The mass of each half becomes \( \frac{M}{2} \) (where \( M \) is the original mass).
- The length of each half becomes \( \frac{L}{2} \) (where \( L \) is the original length).

### Step 4: Calculate the new magnetic moment for each half
The magnetic moment \( M' \) for each half can be calculated as:
\[ M' = \text{pole strength} \times \text{distance between poles} = \frac{M}{2} \times \frac{L}{2} = \frac{ML}{4} \]

### Step 5: Calculate the moment of inertia for each half
The moment of inertia \( I' \) for each half can be calculated using the formula for a thin rod:
\[ I' = \frac{1}{3} m l^2 \]
Substituting \( m = \frac{M}{2} \) and \( l = \frac{L}{2} \):
\[ I' = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L^2}{4}\right) = \frac{ML^2}{24} \]

### Step 6: Substitute the new values into the period formula
Now we substitute \( I' \) and \( M' \) into the period formula:
\[ T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{\frac{ML^2}{24}}{\frac{ML}{4}B}} \]
This simplifies to:
\[ T' = 2\pi \sqrt{\frac{ML^2}{24} \cdot \frac{4}{MLB}} = 2\pi \sqrt{\frac{4L}{24B}} = 2\pi \sqrt{\frac{L}{6B}} \]

### Step 7: Relate the new period to the original period
Since the original period \( T = 2\pi \sqrt{\frac{L}{MB}} \), we can relate the two periods:
\[ T' = \frac{T}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \text{ seconds} \]

### Step 8: Calculate the final result
Given that \( T = 4 \) seconds, we find:
\[ T' = 2 \text{ seconds} \]

Thus, the period of oscillation of each half when freely suspended is **2 seconds**.

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