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A string 120 cm in length sustains a standing wave, with the points of string at which the displacement amplitude is equal to sqrt(2 mm being separated by 5.0 cm, Find the maximum displacement amplitude. Also find the harmonic corresponding to this wave. |
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Answer» Solution :From figure points A, B, C, D and E are having equal displacement amplitude. Further `x_(E)-x_(A)=lambda=4xx15=60 cm` As `lambda=(2l)/(p)=(2xx120)/(p)=60` `:.p=(2xx120)/(60)=4` So, it corresponds to 4th harmonic. Also, distance of node from A is 7.5 cm as distance between B and C is 15 cm and node is between them. TAKING node at origin, the amplitude of stationary wave can be WRITTEN as `a=A sin kx` Here `a= sqrt(mm), k=(2PI)/(lambda)=(2pi)/(60) and x=7.5 cm` `:. sqrt(2)=A sin((2pi)/(60)xx7.5)= A "sin" (pi)/(4)` Hence A=2 mm. 
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