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Questions  

Equation of travelling wave on stretched string of linear density 5g/m is y=0.03sin450t9x where distance and time are measured in S.I units . The tension in the string is

a
10N
b
12.5N
c
7.5N
d
5N

detailed solution

Correct option is B

y=0.03sin450t−9x  Wave Speed : v=ωk=4509=50 m/s  Also, wave speed = v = Tμ=T5×10−3 ⇒50 = T5×10−3 ⇒T=12.5N

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A transverse wave is represented by the equation

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For what value of λ, the maximum particle velocity equal to two times the wave velocity


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