If ` y=alpha cos omega t+b sin omegat`, show that it represents SHM. Determine its amplitude.

Answer

To show that the equation \( y = \alpha \cos(\omega t) + b \sin(\omega t) \) represents simple harmonic motion (SHM) and to determine its amplitude, we can follow these steps:

### Step 1: Rewrite the Equation
The given equation is:
\[ y = \alpha \cos(\omega t) + b \sin(\omega t) \]

### Step 2: Use Trigonometric Identity
We can rewrite this equation in a single sinusoidal form using the trigonometric identity:
\[ R \cos(\omega t - \phi) = R \cos(\phi) \cos(\omega t) + R \sin(\phi) \sin(\omega t) \]
where \( R = \sqrt{\alpha^2 + b^2} \) and \( \tan(\phi) = \frac{b}{\alpha} \).

### Step 3: Identify the Amplitude
From the identity, we can identify that:
\[ R = \sqrt{\alpha^2 + b^2} \]
This \( R \) represents the amplitude of the SHM.

### Step 4: Show that it Satisfies the SHM Condition
To show that this represents SHM, we need to verify that it satisfies the second-order differential equation for SHM:
\[ \frac{d^2y}{dt^2} + \omega^2 y = 0 \]

### Step 5: Calculate the First and Second Derivatives
1. First derivative:
\[ \frac{dy}{dt} = -\alpha \omega \sin(\omega t) + b \omega \cos(\omega t) \]
2. Second derivative:
\[ \frac{d^2y}{dt^2} = -\alpha \omega^2 \cos(\omega t) - b \omega^2 \sin(\omega t) \]
\[ = -\omega^2 (\alpha \cos(\omega t) + b \sin(\omega t)) = -\omega^2 y \]

### Step 6: Substitute into the SHM Equation
Substituting the second derivative back into the SHM equation:
\[ -\omega^2 y + \omega^2 y = 0 \]
This confirms that the equation satisfies the SHM condition.

### Conclusion
Thus, we have shown that the equation \( y = \alpha \cos(\omega t) + b \sin(\omega t) \) represents SHM, and the amplitude of the motion is:
\[ A = \sqrt{\alpha^2 + b^2} \]