The amplitude of a wave represented by displacement equation y=(1/√a) sin ω t ± (1/√b) cos ω t will be

Question

The amplitude of a wave represented by displacement equation y=(1/√a) sin ω t ± (1/√b) cos ω t will be (A) (a+b/a b) (B) (√a+√b/a b) (C) (√a ± √b/a b) (D) √(a+b/a b)

Tardigrade Admin · Accepted Answer

y=(1/√a) sin ω t ± (1/√b) sin (ω t+(π/2)) Here phase difference =(π/2) Resultant amp. =√((1/√a))2+((1/√b))2 =√(1/a)+(1/b)=√(a+b/a b)

Q. The amplitude of a wave represented by displacement equation $y = \frac{1}{a} sin ω t \pm \frac{1}{b} cos ω t$ will be

$\frac{a + b}{ab}$

$\frac{a + b}{ab}$

$\frac{a \pm b}{ab}$

$\frac{a + b}{ab}$

$y = \frac{1}{a} sin ω t \pm \frac{1}{b} sin (ω t + \frac{π}{2})$
Here phase difference $= \frac{π}{2}$
Resultant amp. $= (\frac{1}{a})^{2} + (\frac{1}{b})^{2}$
$= \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$

Q. The amplitude of a wave represented by displacement equation y=a​1​sinωt±b​1​cosωt will be

aba+b​

aba​+b​​

aba​±b​​

aba+b​​

y=a​1​sinωt±b​1​sin(ωt+2π​) Here phase difference =2π​ Resultant amp. =(a​1​)2+(b​1​)2​ =a1​+b1​​=aba+b​​

Q. The amplitude of a wave represented by displacement equation $y = \frac{1}{a} sin ω t \pm \frac{1}{b} cos ω t$ will be

$\frac{a + b}{ab}$

$\frac{a + b}{ab}$

$\frac{a \pm b}{ab}$

$\frac{a + b}{ab}$

$y = \frac{1}{a} sin ω t \pm \frac{1}{b} sin (ω t + \frac{π}{2})$
Here phase difference $= \frac{π}{2}$
Resultant amp. $= (\frac{1}{a})^{2} + (\frac{1}{b})^{2}$
$= \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab}$