Question

On the superposition of two harmonic oscillations represented by ${{x}_{1}}=a\sin \left( \omega t+{{\phi }_{1}} \right)$ and ${{x}_{2}}=a\sin \left( \omega t+{{\phi }_{2}} \right)$, a resulting oscillation with the same time period and amplitude is obtained. The value of ${{\phi }_{1}}-{{\phi }_{2}}$ is
$A)\text{ }{{120}^{0}}$
$B)\text{ 9}{{0}^{0}}$
$C)\text{ 6}{{0}^{0}}$
$D)\text{ }{{15}^{0}}$

Answer 1

This problem can be solved by finding out the expression for the superimposed wave by adding the two given waves mathematically and using trigonometric identities and then solving simultaneously according to the given conditions for the time period and amplitude of the superimposed wave.

Complete answer:
Let us analyze the question.
The representation of one of the harmonic oscillations is ${{x}_{1}}=a\sin \left( \omega t+{{\phi }_{1}} \right)$.
The representation of the second harmonic oscillations is ${{x}_{2}}=a\sin \left( \omega t+{{\phi }_{2}} \right)$.
Let the superimposed harmonic oscillation be $x$.

Now,
$x={{x}_{1}}+{{x}_{2}}$

$\Rightarrow x=a\sin \left( \omega t+{{\phi }_{1}} \right)+a\sin \left( \omega t+{{\phi }_{2}} \right)$
$\Rightarrow x=a\left( \sin \left( \omega t+{{\phi }_{1}} \right)+\sin \left( \omega t+{{\phi }_{2}} \right) \right)$
$\Rightarrow x=a\left( 2\sin \left( \dfrac{\omega t+{{\phi }_{1}}+\omega t+{{\phi }_{2}}}{2} \right)\cos \left( \dfrac{\omega t+{{\phi }_{1}}-\left( \omega t+{{\phi }_{2}} \right)}{2} \right) \right)$ $\because \sin A+\sin B=2\sin \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
$\Rightarrow x=a\left( 2\sin \left( \dfrac{2\omega t+{{\phi }_{1}}+{{\phi }_{2}}}{2} \right)\cos \left( \dfrac{\omega t+{{\phi }_{1}}-\omega t-{{\phi }_{2}}}{2} \right) \right)$
$\Rightarrow x=a\left( 2\sin \left( \omega t+\dfrac{{{\phi }_{1}}+{{\phi }_{2}}}{2} \right)\cos \left( \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2} \right) \right)$
$\Rightarrow x=2a\cos \left( \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2} \right)\left( \sin \left( \omega t+\dfrac{{{\phi }_{1}}+{{\phi }_{2}}}{2} \right) \right)$

Comparing this to the general expression for a harmonic oscillation
$x=A\sin \left( \omega t+\phi \right)$
Where $A$ is the amplitude, $\omega$ is the angular frequency and $\phi$ is the initial phase, we get
$A=2a\cos \left( \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2} \right)$ --(1)

Also, according to the question, the amplitude of this superimposed wave is the same as that of the component oscillations.
$\Rightarrow A=a$ --(2)

Putting (2) in (1), we get
$a=2a\cos \left( \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2} \right)$
$\Rightarrow 1=2\cos \left( \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2} \right)$
$\Rightarrow \dfrac{1}{2}=\cos \left( \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2} \right)$
$\Rightarrow \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2}={{\cos }^{-1}}\left( \dfrac{1}{2} \right)$
$\Rightarrow \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2}={{60}^{0}}$ $\left( \because {{\cos }^{-1}}\left( \dfrac{1}{2} \right)={{60}^{0}} \right)$
$\Rightarrow {{\phi }_{1}}-{{\phi }_{2}}=2\times {{60}^{0}}={{120}^{0}}$
Therefore, we have got the required value for ${{\phi }_{1}}-{{\phi }_{2}}$.

So, the correct answer is “Option A”.

Note:
Students must also note that $\Rightarrow \dfrac{{{\phi }_{1}}-{{\phi }_{2}}}{2}=-{{60}^{0}},{{\phi }_{1}}-{{\phi }_{2}}=-{{120}^{0}}={{360}^{0}}-{{120}^{0}}={{240}^{0}}$
Is also a valid answer. However, since this answer is not given in the options, we do not consider it. Students must also have understood that when two waves get superimposed, the amplitude of the resultant wave can be written in terms of the amplitude of the component waves and the difference in the phase angles of the two component waves.