Question

The amplitude of a wave, represented by displacement equation \( y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t \) will be

• A. \( \frac{a + b}{ab} \)
• B. \( \frac{\sqrt{a} + \sqrt{b}}{ab} \)
• C. \( \frac{\sqrt{a} \pm \sqrt{b}}{ab} \)
• D. \( \sqrt{\frac{a + b}{ab}} \)

Hint:

When two simple harmonic motions of the same frequency are superimposed, the amplitude of the resultant motion is given by the square root of the sum of the squares of the individual amplitudes. If the displacements are \( y_1 = A_1 \sin(\omega t + \phi_1) \) and \( y_2 = A_2 \sin(\omega t + \phi_2) \), the resultant amplitude is \( A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\phi_1 - \phi_2)} \). In this case, rewrite the cosine term as a sine term with a phase shift and then apply the formula. Alternatively, use the form \( A_1 \sin \omega t + A_2 \cos \omega t = R \sin (\omega t + \delta) \), where \( R = \sqrt{A_1^2 + A_2^2} \).

Answer 1

The Correct Option is D

The given displacement equation is \( y = \frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t \).
This equation represents the superposition of two simple harmonic motions with the same angular frequency \( \omega \).
The general form of such a superposition is \( y = A_1 \sin (\omega t) + A_2 \cos (\omega t) \), where \( A_1 = \frac{1}{\sqrt{a}} \) and \( A_2 = \pm \frac{1}{\sqrt{b}} \).
The amplitude \( A \) of the resultant wave is given by \( A = \sqrt{A_1^2 + A_2^2} \).
Substituting the values of \( A_1 \) and \( A_2 \): $$ A = \sqrt{\left(\frac{1}{\sqrt{a}}\right)^2 + \left(\pm \frac{1}{\sqrt{b}}\right)^2} $$ $$ A = \sqrt{\frac{1}{a} + \frac{1}{b}} $$ To simplify the expression under the square root, find a common denominator: $$ A = \sqrt{\frac{b}{ab} + \frac{a}{ab}} $$ $$ A = \sqrt{\frac{a + b}{ab}} $$ The amplitude of the wave is \( \sqrt{\frac{a + b}{ab}} \).