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Q.

A uniform string of length 1.5 m has two successive harmonics of frequencies70 Hz and84 Hz . The speed of the wave in the string (in ms-1 ) is :

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a

84

b

42

c

21

d

10.5

answer is B.

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Detailed Solution

Given , Length l=1.5 m Frequency n1=70 Hz Frequency n2=84 Hz As   λ2=1.5 m ⇒λ=3 m n'=n1−n2 n'=84−70=14Hz ∴v=nλ ⇒14×3=42 m/s
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