A body is executing a linear simple harmonic motion. At t=0, it is at one of the extreme position and travels 3cm in the first second and 4cm in the next second moving towards the other extreme position .
The amplitude of this S.H.M is

Options

A. 17/18cm
B. 16/17cm
C. 19/18cm
D. 18/5cm

Explanation

To solve the problem step by step, we will analyze the motion of the body executing simple harmonic motion (SHM) and derive the amplitude based on the distances traveled in the given time intervals.

### Step 1: Understanding the Motion
At \( t = 0 \), the body is at one extreme position, which we can denote as \( x = A \) (where \( A \) is the amplitude). The body then moves toward the mean position and travels 3 cm in the first second.

### Step 2: Displacement After 1 Second
After 1 second, the displacement can be expressed as:
\[ x(1) = A \cos(\omega \cdot 1) \]
The body travels 3 cm towards the mean position, so:
\[ A - x(1) = 3 \implies A - A \cos(\omega) = 3 \]
This simplifies to:
\[ A(1 - \cos(\omega)) = 3 \quad \text{(Equation 1)} \]

### Step 3: Displacement After 2 Seconds
After 2 seconds, the displacement is:
\[ x(2) = A \cos(\omega \cdot 2) \]
The body travels an additional 4 cm towards the mean position, so:
\[ A - x(2) = 3 + 4 = 7 \implies A - A \cos(2\omega) = 7 \]
This simplifies to:
\[ A(1 - \cos(2\omega)) = 7 \quad \text{(Equation 2)} \]

### Step 4: Using the Cosine Double Angle Formula
Using the double angle formula, we know:
\[ \cos(2\omega) = 2\cos^2(\omega) - 1 \]
Substituting this into Equation 2 gives:
\[ A(1 - (2\cos^2(\omega) - 1)) = 7 \]
This simplifies to:
\[ A(2 - 2\cos^2(\omega)) = 7 \]
or
\[ 2A(1 - \cos^2(\omega)) = 7 \]
Since \( 1 - \cos^2(\omega) = \sin^2(\omega) \), we can write:
\[ 2A \sin^2(\omega) = 7 \quad \text{(Equation 3)} \]

### Step 5: Relating Equations 1 and 3
From Equation 1, we have:
\[ A(1 - \cos(\omega)) = 3 \implies A = \frac{3}{1 - \cos(\omega)} \]
Substituting this into Equation 3:
\[ 2 \left(\frac{3}{1 - \cos(\omega)}\right) \sin^2(\omega) = 7 \]
This simplifies to:
\[ \frac{6 \sin^2(\omega)}{1 - \cos(\omega)} = 7 \]
Cross-multiplying gives:
\[ 6 \sin^2(\omega) = 7(1 - \cos(\omega)) \]

### Step 6: Using the Identity \( \sin^2(\omega) = 1 - \cos^2(\omega) \)
We can express \( \sin^2(\omega) \) in terms of \( \cos(\omega) \):
\[ 6(1 - \cos^2(\omega)) = 7(1 - \cos(\omega)) \]
Expanding and rearranging gives:
\[ 6 - 6\cos^2(\omega) = 7 - 7\cos(\omega) \]
Rearranging leads to:
\[ 6\cos^2(\omega) - 7\cos(\omega) + 1 = 0 \]

### Step 7: Solving the Quadratic Equation
Using the quadratic formula \( \cos(\omega) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ \cos(\omega) = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 6 \cdot 1}}{2 \cdot 6} \]
\[ = \frac{7 \pm \sqrt{49 - 24}}{12} \]
\[ = \frac{7 \pm 5}{12} \]
Calculating the two possible values:
1. \( \cos(\omega) = 1 \) (not possible as it implies no motion)
2. \( \cos(\omega) = \frac{1}{6} \)

### Step 8: Finding the Amplitude
Substituting \( \cos(\omega) = \frac{1}{6} \) back into Equation 1:
\[ A(1 - \frac{1}{6}) = 3 \implies A \cdot \frac{5}{6} = 3 \implies A = \frac{3 \cdot 6}{5} = \frac{18}{5} \text{ cm} \]

### Final Answer
The amplitude of the SHM is:
\[ \boxed{\frac{18}{5} \text{ cm}} \]