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PhysicsWave on StringMiscellaneousHard2 minPYQ_2014
PhysicsHardsingle choice

The total length of a sonometer wire fixed between two bridges is110 cm. Now, two more bridges are placed to divide the length of the wire in the ratio6:3:2. If the tension in the wire is400 Nand the mass per unit length of the wire is0.01 kg m-1, then the minimum common frequency with which all the three parts can vibrate, is

Options:

Answer:
A
Solution:

l 1 : l 2 : l 3 =6:3:2

The frequancy in any nth mode for a segment is f=nv2l=constant

no of loops ∝ length of the segment

so, no of loops are in the ratio 6:3:2

hence total loops =11

The string is divided in 60 cm, 30 cm, & 20 cm part such that for minimum frequency, the wavelength is maximum

λ2=111x110=10cm

f = V λ = 1 λ · F μ = 1 0.2 4 0 0 0.01 = 1 0 0 0 Hz

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Miscellaneous
2mℹ️ Source: PYQ_2014

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