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 direction. 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 solve the problem of finding the amplitude of a particle performing Simple Harmonic Motion (SHM) given that it travels a distance \( a \) in the first second and a distance \( b \) in the second second, we can follow these steps:

### Step 1: Understand the Motion
The particle starts from rest and travels a distance \( a \) in the first second and \( b \) in the second second. This indicates that the distances covered in each second are not equal, which is a characteristic of SHM.

### Step 2: Use the Displacement Equation
The displacement of a particle in SHM can be represented as:
\[ x(t) = A \cos(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

### Step 3: Analyze the Distances
1. In the first second, the particle travels a distance \( a \):
\[ x(1) = A \cos(\omega \cdot 1) \]
The distance traveled in the first second is:
\[ a = x(1) - x(0) = A \cos(\omega \cdot 1) - 0 \]
Therefore, we have:
\[ A \cos(\omega) = a \]

2. In the second second, the particle travels a distance \( b \):
\[ x(2) = A \cos(\omega \cdot 2) \]
The distance traveled in the second second is:
\[ b = x(2) - x(1) = A \cos(\omega \cdot 2) - A \cos(\omega \cdot 1) \]
Thus, we can write:
\[ b = A \cos(\omega \cdot 2) - a \]

### Step 4: Substitute and Rearrange
From the first equation, we can express \( A \cos(\omega) \) as:
\[ A \cos(\omega) = a \]
Substituting this into the second equation gives us:
\[ b = A \cos(\omega \cdot 2) - a \]

Now, we can express \( A \cos(\omega \cdot 2) \) in terms of \( a \) and \( b \):
\[ A \cos(\omega \cdot 2) = b + a \]

### Step 5: Use the Cosine Double Angle Identity
Using the identity \( \cos(2\theta) = 2\cos^2(\theta) - 1 \), we can relate \( A \cos(\omega) \) and \( A \cos(\omega \cdot 2) \):
\[ A \cos(\omega \cdot 2) = 2(A \cos(\omega))^2/A - 1 \]
Substituting \( A \cos(\omega) = a \):
\[ A \cos(\omega \cdot 2) = 2\left(\frac{a}{A}\right)^2A - 1 \]

### Step 6: Solve for Amplitude \( A \)
Now, we can equate the two expressions for \( A \cos(\omega \cdot 2) \):
\[ b + a = 2\left(\frac{a}{A}\right)^2A - 1 \]

Rearranging gives us a quadratic equation in terms of \( A \):
\[ 2a^2 - A(b + a + 1) = 0 \]

### Step 7: Final Expression for Amplitude
Solving this quadratic equation will yield the amplitude \( A \):
\[ A = \frac{2a^2}{b + a + 1} \]

### Conclusion
Thus, the amplitude of the SHM is given by:
\[ A = \frac{2a^2}{b + a + 1} \]