(a) The motion of the particle in simple harmonic motion is given by `x = a sin omega t`. If its speed is `u`, when the displacement is `x_(1)` and speed is `v`, when the displacement is `x_(2)`, show that the amplitude of the motion is
`A = [(v^(2)x_(1)^(2) - u^(2)x_(2)^(2))/(v^(2) - u^(2))]^(1//2)`
(b) A particle is moving with simple harmonic motion is a straight line. When the distance of the particle from the equilibrium position has the values ` x_(1)` and `x_(2)` the corresponding values of velocity are `u_(1)` and `u_(2)`, show that the period is
`T = 2pi[(x_(2)^(2) - x_(1)^(2))/(u_(1)^(2) - u_(2)^(2))]^(1//2)`

Explanation

To solve the given problem step by step, we will break it down into two parts as per the question.

### Part (a)

**Step 1: Start with the equation of motion for SHM.**
The motion of a particle in simple harmonic motion (SHM) is given by:
\[ x = A \sin(\omega t) \]
where \(A\) is the amplitude, \(x\) is the displacement, and \(\omega\) is the angular frequency.

**Step 2: Find the expression for velocity.**
The velocity \(v\) of the particle can be found by differentiating the displacement with respect to time:
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t) \]

**Step 3: Relate velocity to displacement.**
From the equation of motion, we can express the amplitude in terms of velocity and displacement:
\[ v^2 = A^2 \omega^2 \cos^2(\omega t) \]
Using the identity \(\cos^2(\theta) = 1 - \sin^2(\theta)\), we can rewrite this as:
\[ v^2 = A^2 \omega^2 \left(1 - \left(\frac{x}{A}\right)^2\right) \]
Rearranging gives:
\[ v^2 = A^2 \omega^2 - \omega^2 x^2 \]
Thus,
\[ A^2 = \frac{v^2 + \omega^2 x^2}{\omega^2} \]

**Step 4: Apply this for two different displacements.**
For displacement \(x_1\) with speed \(u\):
\[ u^2 = A^2 \omega^2 - \omega^2 x_1^2 \]
For displacement \(x_2\) with speed \(v\):
\[ v^2 = A^2 \omega^2 - \omega^2 x_2^2 \]

**Step 5: Solve these equations to find \(A\).**
From both equations, we can express \(A^2\):
1. From \(u^2\):
\[ A^2 = \frac{u^2 + \omega^2 x_1^2}{\omega^2} \]
2. From \(v^2\):
\[ A^2 = \frac{v^2 + \omega^2 x_2^2}{\omega^2} \]

Equating both expressions for \(A^2\):
\[ \frac{u^2 + \omega^2 x_1^2}{\omega^2} = \frac{v^2 + \omega^2 x_2^2}{\omega^2} \]
This simplifies to:
\[ u^2 + \omega^2 x_1^2 = v^2 + \omega^2 x_2^2 \]

**Step 6: Isolate \(A\).**
Rearranging gives:
\[ \omega^2 (x_1^2 - x_2^2) = v^2 - u^2 \]
Thus, we can express \(A\) as:
\[ A = \sqrt{\frac{v^2 x_1^2 - u^2 x_2^2}{v^2 - u^2}} \]

This completes the proof for part (a).

### Part (b)

**Step 1: Use the velocity expressions for SHM.**
From the previous part, we have:
\[ u_1 = \omega \sqrt{A^2 - x_1^2} \]
\[ u_2 = \omega \sqrt{A^2 - x_2^2} \]

**Step 2: Square both equations.**
Squaring both sides gives:
\[ u_1^2 = \omega^2 (A^2 - x_1^2) \]
\[ u_2^2 = \omega^2 (A^2 - x_2^2) \]

**Step 3: Rearrange to find \(A^2\).**
From these equations, we can express \(A^2\):
1. From \(u_1^2\):
\[ A^2 = \frac{u_1^2}{\omega^2} + x_1^2 \]
2. From \(u_2^2\):
\[ A^2 = \frac{u_2^2}{\omega^2} + x_2^2 \]

**Step 4: Set the equations equal to each other.**
Equating both expressions for \(A^2\):
\[ \frac{u_1^2}{\omega^2} + x_1^2 = \frac{u_2^2}{\omega^2} + x_2^2 \]

**Step 5: Solve for \(\omega^2\).**
Rearranging gives:
\[ \frac{u_1^2 - u_2^2}{\omega^2} = x_2^2 - x_1^2 \]
Thus,
\[ \omega^2 = \frac{u_1^2 - u_2^2}{x_2^2 - x_1^2} \]

**Step 6: Find the period \(T\).**
The period \(T\) is related to angular frequency \(\omega\) by:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \(\omega\):
\[ T = 2\pi \sqrt{\frac{x_2^2 - x_1^2}{u_1^2 - u_2^2}} \]

This completes the proof for part (b).