Question

The differential equation obtained by eliminating A and B from $y = A \cos \omega t + B \sin \omega t$ is

• A. $\frac{d^2y}{dt^2} + \omega^2 y = 0$
• B. $\frac{d^2y}{dt^2} + \omega y^2 = 0$
• C. $\frac{d^2y}{dt^2} - \omega^2 y = 0$
• D. $\frac{d^2y}{dt^2} - \omega y^2 = 0$

Hint:

Any linear combination of sine and cosine functions sharing the exact same angular frequency parameter $\omega$ (e.g., $y = A\cos\omega t + B\sin\omega t$) is a solution to the classical harmonic oscillator differential equation $\ddot{y} + \omega^2 y = 0$. Recognizing this physical standard lets you pick option (A) instantly!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires us to formulate a differential equation by eliminating the two arbitrary constants $A$ and $B$ from the given equation representing simple harmonic motion.

Step 2: Key Formula or Approach:
Since there are exactly two arbitrary constants ($A$ and $B$), we must differentiate the function twice with respect to the independent variable $t$, and then algebraically substitute the original equation back to eliminate the parameters.

Step 3: Detailed Explanation:
The given equation is:
$$y = A \cos \omega t + B \sin \omega t$$ Differentiating once with respect to $t$ using the chain rule:
$$\frac{dy}{dt} = -A\omega \sin \omega t + B\omega \cos \omega t$$ Differentiating a second time with respect to $t$:
$$\frac{d^2y}{dt^2} = -A\omega^2 \cos \omega t - B\omega^2 \sin \omega t$$ Factor out the common multiplier $-\omega^2$ from both terms on the right-hand side:
$$\frac{d^2y}{dt^2} = -\omega^2 (A \cos \omega t + B \sin \omega t)$$ Notice that the expression inside the parentheses is exactly our original function $y$. Substituting $y$ back into the expression yields:
$$\frac{d^2y}{dt^2} = -\omega^2 y$$ Rearranging all terms to the left-hand side gives the standard linear differential equation:
$$\frac{d^2y}{dt^2} + \omega^2 y = 0$$ This matches option (A).

Step 4: Final Answer:
The resulting differential equation is $\frac{d^2y}{dt^2} + \omega^2 y = 0$, which corresponds to option (A).