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 solve the problem step by step, we will follow the principles of Simple Harmonic Motion (SHM).

### Step 1: Determine the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Where \(T\) is the period of the motion. Given that \(T = 20\) seconds, we can calculate ω:
\[ \omega = \frac{2\pi}{20} = \frac{\pi}{10} \text{ radians per second} \]

### Step 2: Write the equation for velocity in SHM
The velocity \(v\) in SHM can be expressed as:
\[ v = A \omega \cos(\omega t) \]
Where \(A\) is the amplitude, and \(t\) is the time after passing the mean position.

### Step 3: Substitute the known values
We know that \(v = 5 \, \text{cm/s}\) and \(t = 2 \, \text{s}\). We can substitute these values into the velocity equation:
\[ 5 = A \left(\frac{\pi}{10}\right) \cos\left(\frac{\pi}{10} \cdot 2\right) \]

### Step 4: Calculate \(\cos(\omega t)\)
First, we calculate \(\omega t\):
\[ \omega t = \frac{\pi}{10} \cdot 2 = \frac{2\pi}{10} = \frac{\pi}{5} \]
Now, we can find \(\cos\left(\frac{\pi}{5}\right)\). From the problem, we know that:
\[ \cos(36^\circ) = 0.809 \]
Thus:
\[ \cos\left(\frac{\pi}{5}\right) = 0.809 \]

### Step 5: Substitute \(\cos(\omega t)\) back into the equation
Now we substitute \(\cos\left(\frac{\pi}{5}\right)\) back into the velocity equation:
\[ 5 = A \left(\frac{\pi}{10}\right) (0.809) \]

### Step 6: Solve for the amplitude \(A\)
Rearranging the equation to solve for \(A\):
\[ A = \frac{5}{\left(\frac{\pi}{10}\right) (0.809)} \]
This simplifies to:
\[ A = \frac{5 \cdot 10}{\pi \cdot 0.809} \]
\[ A = \frac{50}{\pi \cdot 0.809} \]

### Step 7: Calculate the numerical value of \(A\)
Using \(\pi \approx 3.14\):
\[ A \approx \frac{50}{3.14 \cdot 0.809} \approx \frac{50}{2.54} \approx 19.69 \, \text{cm} \]

### Final Answer
The amplitude of the bob is approximately:
\[ A \approx 19.67 \, \text{cm} \]