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PhysicsSHMMiscellaneousEasy2 minQB
PhysicsEasynumerical

A simple harmonic wave of amplitude 8 unit travels along positive x-axis. At a given instant of time, for a particle at a distance of 10 cm from the origin, the displacement is + 6 unit and for a particle at a distance of 25 cm from the origin, the displacement is + 4 unit. What is the wavelength (in metre) of the wave?

Answer:
0.25
Solution:

Y=Asin⁡2πλvt-x

YA=sin⁡2πtT-xλ

In first case,Y1A=sin⁡2πtT-x1λ

Here,Y1=+ 6,A=8,x1=10 cm

68=sin⁡2πtT-10λ.... (i)

Similarly in the second case,

48=sin⁡2πtT-25λ... (ii)

From equation (i),

2π​ tT-10λ=sin-1⁡68=0.85 rad

⇒  tT-10λ=0.14.... (iii)

Similarly from equation (ii),

2πtT-25λ=sin-1⁡48=π6 rad

⇒       tT=25λ+​0.08.... (iv)

Subtracting equation (iv) from equation (iii), we get

  15λ=0.06

⇒              λ=0.25 m

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Miscellaneous
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