Question

The amplitude of a wave represented by displacement equation
$\sf y = \dfrac{1}{ \sqrt{a} } \: sin \omega t ± \dfrac{1}{ \sqrt{b} } cos \omega t$
will be
$\sf(1) \: \dfrac{a + b}{ab}$
$\sf(2) \: \dfrac{ \sqrt{a} + \sqrt{b} }{ab}$
$\sf(3) \: \dfrac{ \sqrt{a} ± \sqrt{b} }{ab}$
$\sf(4) \: \sqrt{ \dfrac{a + b}{ab} }$

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Answer 1

Given :

The amplitude of a wave represented by displacement equation $\sf y = \dfrac{1}{ \sqrt{a} } \: sin \omega t \pm \dfrac{1}{ \sqrt{b} } cos \omega t$

To find :

Amplitude of the wave

Solution :

given that,

$\sf y = \dfrac{1}{ \sqrt{a} } \: sin \omega t ± \dfrac{1}{ \sqrt{b} } cos \omega t$

here we have two waves, If two or more waves reach a particle simultaneously the resultant displacement of the particle in the medium is the vectorial sum of displacement due to individual waves.

$\sf y = y_1 + y_2 \: ....$

According to the question,

$\sf y_1 = \dfrac{1}{ \sqrt{a} } \: sin \omega t$

$\sf y_2 = \dfrac{1}{ \sqrt{b} } cos \omega t$

we know that,

If two vector y₁and y₂ of magnitude y₁and y₂ are acting at an angle θ then the magnitude of their resultant using parallelogram method of vector addition is

$\sf R = \sqrt{ y_1 + y_2 + 2y_1y_2cos \theta}$

By substituting all the values,

$\sf A = \sqrt{ \bigg( \dfrac{1}{ \sqrt{a} }\bigg) ^{2} +\bigg( \dfrac{1}{ \sqrt{b} }\bigg) ^{2} + 2 \times \dfrac{1}{ \sqrt{a} } \times \dfrac{1}{ \sqrt{b} } cos (90)}$

$\sf A = \sqrt{ \dfrac{1}{ a } + \dfrac{1}{ b }+ 2 \times \dfrac{1}{ \sqrt{a} } \times \dfrac{1}{ \sqrt{b} } (0)}$

$\sf A = \sqrt{ \dfrac{1}{ a } + \dfrac{1}{ b }}$

$\sf A = \sqrt{ \dfrac{b + a}{ ab } }$

thus, option (d) is correct.