Question

# A string, fixed at both ends, vibrates in a resonant node with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. find the length of the string.

A

18 cm

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B

8 cm

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C

10 cm

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D

cannot be determined

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Solution

## The correct option is B 8 cm In the first case the distance between cause cu tine nodes is 2 cm, Which implies λ2 = 2cm cm or λ = 4cm If v is the velocity of wave on the string Frequency,f = f = vλ = v4- - - - -- (1) When the distance between the nodes become 1.6 cm λ′2 = 1.6cm ⇒ λ′=3.2cm Frequency,f′ = vλ′ = v3.2 - - - - - (2) Also for some n,f = nv2L ⇒ v4 = nv2L [from(1)] n = L2 - - - - - (3) then for some n+1,f′ = (n+1)v2L ⇒ v3.2 = (n+1)v2L [from(2)] 2L3.2 = n + 1 ----------- (4) From (3) and (4) 2L3.2 = L2 + 1 4L = 3.2 (L+2) 4L = 3.2L + 6.4 0.8L = 6.4 L = 8cm

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