A particle performs SHM in a straight line. In the first second, starting from rest, it travels a distance a and in the next second it travels a distance b in the same side of mean position. The amplitude of the SHM is

Options

A. a-b
B. `(2a-b)/3`
C. `(2a^(2))/(3a-b)`
D. None of these

Answer

To find the amplitude of the simple harmonic motion (SHM) of the particle, we can follow these steps:

### Step 1: Understand the motion of the particle
The particle starts from rest and travels a distance \( a \) in the first second and a distance \( b \) in the second second in the same direction from the mean position.

### Step 2: Write the equations for displacement
Since the particle is performing SHM, we can express the displacement \( x \) at any time \( t \) as:
\[ x(t) = A \cos(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

### Step 3: Analyze the first second
At \( t = 1 \) second, the particle has traveled a distance \( a \). Therefore, the displacement from the mean position can be expressed as:
\[ x(1) = A \cos(\omega \cdot 1) = A \cos(\omega) \]
Given that the particle starts from rest and travels a distance \( a \), we can write:
\[ x(1) = A - a \]
Thus, we have:
\[ A \cos(\omega) = A - a \quad \text{(Equation 1)} \]

### Step 4: Analyze the second second
At \( t = 2 \) seconds, the particle has traveled an additional distance \( b \), so the total distance from the mean position is:
\[ x(2) = A - a - b \]
Using the SHM displacement equation, we can write:
\[ x(2) = A \cos(2\omega) \]
Thus, we have:
\[ A \cos(2\omega) = A - a - b \quad \text{(Equation 2)} \]

### Step 5: Use trigonometric identities
We can use the double angle identity for cosine:
\[ \cos(2\omega) = 2\cos^2(\omega) - 1 \]
Substituting this into Equation 2 gives:
\[ A (2\cos^2(\omega) - 1) = A - a - b \]

### Step 6: Substitute \( \cos(\omega) \) from Equation 1
From Equation 1, we can express \( \cos(\omega) \):
\[ \cos(\omega) = \frac{A - a}{A} \]
Substituting this into the expression for \( \cos^2(\omega) \):
\[ \cos^2(\omega) = \left(\frac{A - a}{A}\right)^2 \]

### Step 7: Substitute back into the equation
Now substituting \( \cos^2(\omega) \) into the equation:
\[ A \left(2\left(\frac{A - a}{A}\right)^2 - 1\right) = A - a - b \]
This simplifies to:
\[ 2(A - a)^2 - A^2 = A - a - b \]

### Step 8: Solve for \( A \)
Expanding and rearranging the equation:
\[ 2(A^2 - 2aA + a^2) - A^2 = A - a - b \]
\[ 2A^2 - 4aA + 2a^2 - A^2 = A - a - b \]
\[ A^2 - 4aA + 2a^2 = A - a - b \]
Rearranging gives:
\[ A^2 - (4a + 1)A + (2a^2 + a + b) = 0 \]

### Step 9: Use the quadratic formula
Using the quadratic formula \( A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( A \):
\[ A = \frac{(4a + 1) \pm \sqrt{(4a + 1)^2 - 4(1)(2a^2 + a + b)}}{2} \]

### Final Answer
The amplitude \( A \) can be calculated using the above formula, and the specific values of \( a \) and \( b \) will yield the numerical amplitude.

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