A body executes S.H.M. of period 20 seconds. Its velocity is `5 cm s^(-1)`, 2 seconds after it has passed the mean position. Find the amplitude of the bob
(`cos 36^(@) = 0. 809`)

Options

A. 21.45 cm
B. 16. 56 cm
C. 19.67 cm
D. 15. 34 cm

Answer

To find the amplitude of a body executing Simple Harmonic Motion (S.H.M.) with a given period and velocity, we can follow these steps:

### Step 1: Identify the given data
- Period (T) = 20 seconds
- Velocity (V) = 5 cm/s
- Time after passing the mean position (t) = 2 seconds

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{20} = \frac{\pi}{10} \text{ rad/s} \]

### Step 3: Write the expression for velocity in S.H.M.
The velocity (V) in S.H.M. can be expressed as:
\[ V = A \omega \cos(\omega t) \]
where A is the amplitude.

### Step 4: Substitute the known values into the velocity equation
We know V = 5 cm/s and ω = \(\frac{\pi}{10}\) rad/s, and t = 2 seconds. Substitute these values into the equation:
\[ 5 = A \left(\frac{\pi}{10}\right) \cos\left(\frac{\pi}{10} \times 2\right) \]
This simplifies to:
\[ 5 = A \left(\frac{\pi}{10}\right) \cos\left(\frac{\pi}{5}\right) \]

### Step 5: Calculate \(\cos\left(\frac{\pi}{5}\right)\)
Using the given value:
\[ \cos\left(\frac{\pi}{5}\right) = \cos(36^\circ) = 0.809 \]

### Step 6: Substitute \(\cos\left(\frac{\pi}{5}\right)\) into the equation
Now substitute this value back into the equation:
\[ 5 = A \left(\frac{\pi}{10}\right) (0.809) \]

### Step 7: Solve for amplitude (A)
Rearranging the equation to solve for A:
\[ A = \frac{5 \times 10}{\pi \times 0.809} \]
Calculating this gives:
\[ A = \frac{50}{\pi \times 0.809} \]
Using \(\pi \approx 3.14\):
\[ A \approx \frac{50}{3.14 \times 0.809} \approx \frac{50}{2.54} \approx 19.67 \text{ cm} \]

### Final Answer
The amplitude of the bob is approximately **19.67 cm**.

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