The initial position and velocity of a body moving in SHM with period `T = 0.25s` are `x = 5.0cm` and `v = 218cm//s`. What are the amplitude and phase constant of the motion ?

Explanation

To solve the problem of finding the amplitude and phase constant of a body moving in Simple Harmonic Motion (SHM) with given initial conditions, we can follow these steps:

### Step 1: Calculate Angular Frequency (ω)
The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Given that the period \( T = 0.25 \, \text{s} \):
\[ \omega = \frac{2\pi}{0.25} = 8\pi \, \text{rad/s} \]

### Step 2: Set Up the Equations
The general equations for position \( x \) and velocity \( v \) in SHM are:
\[ x = A \sin(\omega t + \phi) \]
\[ v = \omega A \cos(\omega t + \phi) \]
At time \( t = 0 \), we can simplify these equations to:
\[ x(0) = A \sin(\phi) \quad \text{(1)} \]
\[ v(0) = \omega A \cos(\phi) \quad \text{(2)} \]

### Step 3: Substitute Known Values
From the problem, we know:
- \( x(0) = 5.0 \, \text{cm} \)
- \( v(0) = 218 \, \text{cm/s} \)
- \( \omega = 8\pi \, \text{rad/s} \)

Substituting these values into equations (1) and (2):
1. \( 5.0 = A \sin(\phi) \) (Equation 1)
2. \( 218 = 8\pi A \cos(\phi) \) (Equation 2)

### Step 4: Solve the Equations
From Equation (1):
\[ A \sin(\phi) = 5.0 \quad \Rightarrow \quad A = \frac{5.0}{\sin(\phi)} \quad \text{(3)} \]

Substituting Equation (3) into Equation (2):
\[ 218 = 8\pi \left(\frac{5.0}{\sin(\phi)}\right) \cos(\phi) \]
\[ 218 \sin(\phi) = 40\pi \cos(\phi) \]
Dividing both sides by \( \cos(\phi) \):
\[ \frac{218 \sin(\phi)}{\cos(\phi)} = 40\pi \]
This simplifies to:
\[ 218 \tan(\phi) = 40\pi \]
\[ \tan(\phi) = \frac{40\pi}{218} \]

### Step 5: Calculate Phase Constant (φ)
Now we can find \( \phi \):
\[ \phi = \tan^{-1}\left(\frac{40\pi}{218}\right) \]
Calculating this gives:
\[ \phi \approx \frac{\pi}{6} \text{ or } 30^\circ \]

### Step 6: Calculate Amplitude (A)
Now substituting \( \phi \) back into Equation (3) to find \( A \):
\[ A = \frac{5.0}{\sin\left(\frac{\pi}{6}\right)} = \frac{5.0}{0.5} = 10.0 \, \text{cm} \]

### Final Answers
- Amplitude \( A \approx 10.0 \, \text{cm} \)
- Phase constant \( \phi \approx \frac{\pi}{6} \text{ or } 30^\circ \)