A particle moves with simple harmonic motion in a straight line. In first `taus`, after starting form rest it travels a destance a, and in next `tau s` it travels 2a, in same direction, then:

Options

A. (a)amplitude of motion is `3a`.
B. (b) time period of oscllations is `8tau`.
C. (c ) amplitude of motion is `4tau`.
D. (d) time period of oscillations is `6tau`.

Explanation

To solve the problem step by step, we will analyze the motion of the particle undergoing simple harmonic motion (SHM) and derive the necessary parameters.

### Step 1: Understand the Motion
The particle starts from rest, which means it is at the extreme position of its motion. The distance traveled in the first time interval \( \tau \) is \( a \), and in the next time interval \( \tau \), it travels \( 2a \).

### Step 2: Set Up the Equations
1. **First Interval (0 to \( \tau \))**:
- Initial position: \( x_0 = A \) (where \( A \) is the amplitude).
- Distance traveled: \( a \).
- Final position after the first interval:
\[ x_1 = A - a \]

2. **Second Interval (\( \tau \) to \( 2\tau \))**:
- Distance traveled in this interval: \( 2a \).
- Final position after the second interval:
\[ x_2 = x_1 - 2a = (A - a) - 2a = A - 3a \]

### Step 3: Use the SHM Equation
In SHM, the position \( x \) as a function of time \( t \) is given by:
\[ x(t) = A \cos(\omega t) \]
where \( \omega \) is the angular frequency.

1. For the first interval:
\[ A - a = A \cos(\omega \tau) \]
Rearranging gives:
\[ \cos(\omega \tau) = 1 - \frac{a}{A} \quad \text{(Equation 1)} \]

2. For the second interval:
\[ A - 3a = A \cos(2\omega \tau) \]
Rearranging gives:
\[ \cos(2\omega \tau) = 1 - \frac{3a}{A} \quad \text{(Equation 2)} \]

### Step 4: Use the Double Angle Formula
Using the double angle formula for cosine:
\[ \cos(2\theta) = 2\cos^2(\theta) - 1 \]
we can substitute \( \cos(\omega \tau) \) from Equation 1 into Equation 2.

Substituting:
\[ 1 - \frac{3a}{A} = 2\left(1 - \frac{a}{A}\right)^2 - 1 \]

### Step 5: Solve for \( a/A \)
Expanding the right side:
\[ 1 - \frac{3a}{A} = 2\left(1 - \frac{2a}{A} + \frac{a^2}{A^2}\right) - 1 \]
\[ 1 - \frac{3a}{A} = 2 - 4\frac{a}{A} + 2\frac{a^2}{A^2} - 1 \]
\[ 1 - \frac{3a}{A} = 1 - 4\frac{a}{A} + 2\frac{a^2}{A^2} \]
Setting the equation:
\[ -\frac{3a}{A} = -4\frac{a}{A} + 2\frac{a^2}{A^2} \]
Multiplying through by \( A^2 \):
\[ -3aA = -4aA + 2a^2 \]
Rearranging gives:
\[ 2a^2 + aA = 0 \]
Factoring out \( a \):
\[ a(2a + A) = 0 \]
Thus, \( a = 0 \) or \( 2a + A = 0 \). Since \( a \) cannot be zero, we have:
\[ A = -2a \]

### Step 6: Find the Time Period
The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
From the equations derived, we can find the time period \( T \) based on the relationship established.

### Final Result
The amplitude \( A \) is \( 2a \) and the time period \( T \) can be derived from the relationship of \( \omega \).