A particles is SHM is described by the displacement function `x(t)= a cos (omegat+theta).` If the initinal `(t=0)` position of the particle is 1 cm and its initinal velcotiy is `pi cm//s.` The angular frequncy of the particle is `pi rad//s`. Then is amplitude is

Options

A. 1 cm
B. `sqrt2 cm `
C. `2 cm`
D. `2.5 cm`

Answer

To solve the problem, we start with the given displacement function of the particle in simple harmonic motion (SHM):

\[ x(t) = a \cos(\omega t + \theta) \]

### Step 1: Substitute Initial Conditions
At \( t = 0 \), the displacement \( x(0) \) is given as 1 cm. Therefore, we can write:

\[ x(0) = a \cos(\theta) = 1 \]

This gives us our first equation:

\[ a \cos(\theta) = 1 \quad \text{(Equation 1)} \]

### Step 2: Find the Velocity Function
The velocity \( v(t) \) of the particle is the derivative of the displacement function:

\[ v(t) = \frac{dx}{dt} = -a \omega \sin(\omega t + \theta) \]

### Step 3: Substitute Initial Velocity
At \( t = 0 \), the initial velocity \( v(0) \) is given as \( \pi \) cm/s. Therefore, we can write:

\[ v(0) = -a \omega \sin(\theta) = \pi \]

Substituting the value of \( \omega = \pi \) rad/s, we have:

\[ -a \pi \sin(\theta) = \pi \]

This simplifies to:

\[ -a \sin(\theta) = 1 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations:

1. \( a \cos(\theta) = 1 \)
2. \( -a \sin(\theta) = 1 \)

From Equation 1, we can express \( \cos(\theta) \):

\[ \cos(\theta) = \frac{1}{a} \]

From Equation 2, we can express \( \sin(\theta) \):

\[ \sin(\theta) = -\frac{1}{a} \]

### Step 5: Use the Pythagorean Identity
We know that:

\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]

Substituting the expressions for \( \sin(\theta) \) and \( \cos(\theta) \):

\[ \left(-\frac{1}{a}\right)^2 + \left(\frac{1}{a}\right)^2 = 1 \]

This simplifies to:

\[ \frac{1}{a^2} + \frac{1}{a^2} = 1 \]

\[ \frac{2}{a^2} = 1 \]

### Step 6: Solve for Amplitude \( a \)
Multiplying both sides by \( a^2 \):

\[ 2 = a^2 \]

Taking the square root:

\[ a = \sqrt{2} \text{ cm} \]

### Conclusion
The amplitude of the particle is \( \sqrt{2} \) cm.

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