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Question: At an instance of time O is a point on the string at which its displacement is zero. At the same ins...

At an instance of time O is a point on the string at which its displacement is zero. At the same instant a point, P which is at a distance λ8\dfrac{\lambda }{8} from O, has the displacement A. The displacement of a point at a distance λ4\dfrac{\lambda }{4} from O is nA2\dfrac{{nA}}{{\sqrt 2 }}, calculate the value of n?

Explanation

Solution

The shape of the wave (at any fixed instant) as a function of time x is a sine wave. This question can be solved using the wave equation formula. Wavelength λ\lambda is the distance between two consecutive crests or troughs. At point ‘O’, since its displacement is zero, its amplitude is zero; we can consider it as the origin point in the diagram.

Complete step-by-step solution:
At a fixed time t, displacement y varies as a function of position x. That is y=Asin(kx)y = A'\sin (kx), where k= 2πλ\dfrac{{2\pi }}{\lambda }
So from the equation y=Asin(kx)y = A'\sin (kx)
Substituting the value of k we get, y=Asin(2πλx)y = A'\sin (\dfrac{{2\pi }}{\lambda }x)
For point P which is at a distance λ8\dfrac{\lambda }{8} from O, has the displacement A, the wave equation becomes
A=Asin(2πλλ8)A = A'\sin (\dfrac{{2\pi }}{\lambda }*\dfrac{\lambda }{8}) , displacementy=Ay = A; distance x=x = λ8\dfrac{\lambda }{8}
Simplifying we get, A=Asin(π4)A = A'\sin (\dfrac{\pi }{4})
Since sin(π4)=12\sin (\dfrac{\pi }{4}) = \dfrac{1}{\sqrt 2} We get, A=A12A = A'*\dfrac{1}{{\sqrt 2 }}
Therefore by rearranging we get, A=2AA' = \sqrt 2 A………..(1)
For the point at a distance λ4\dfrac{\lambda }{4} from O with displacement nA2\dfrac{{nA}}{{\sqrt 2 }}, the wave equation becomes
y=Asin(2πλx)y = A'\sin (\dfrac{{2\pi }}{\lambda }x)
Substituting y = $$$\dfrac{{nA}}{{\sqrt 2 }}$; x = $$$\dfrac{\lambda }{4}Thewaveequationbecomes, The wave equation becomes,\dfrac{{nA}}{{\sqrt 2 }} = A'\sin (\dfrac{{2\pi }}{\lambda }*\dfrac{\lambda }{4})Simplifyingweget, Simplifying we get,\dfrac{{nA}}{{\sqrt 2 }} = A'\sin (\dfrac{\pi }{2})Since Since\sin (\dfrac{\pi }{2}) = 1,waveequationbecomes, wave equation becomes\dfrac{{nA}}{{\sqrt 2 }} = A'Fromequation(1)weknowthat From equation (1) we know thatA' = \sqrt 2 ASubstitutingforAintheaboveequation Substituting for A’ in the above equation\dfrac{{nA}}{{\sqrt 2 }} = A',gives, gives \dfrac{{nA}}{{\sqrt 2 }} = \sqrt 2 ACrossmultiplyingandcancellingsimilartermsweget, Cross multiplying and cancelling similar terms we get,nA = (\sqrt 2 *\sqrt 2 )A = 2A n = 2$

Note: Wave is a form of disturbance that travels through a medium as a result of the periodic motion of the particles in the medium. Waves not only transport energy but its pattern of disturbance contains information that propagates from one point to another. There are mainly three types of waves: mechanical, electromagnetic and matter waves.