The frequency of the fundamental mode of transverse vibration of stret...
Explanation:
When a wire is stretched and fixed at both ends, it vibrates transversely in its fundamental mode with a certain frequency. The frequency of vibration depends on the length, tension, and mass per unit length of the wire.
Given:
Length of wire, l1 = 1000mm
Frequency of fundamental mode, f1 = 200Hz
Shortened length of wire, l2 = 500mm
Frequency of fundamental mode when shortened, f2 = ?
Solution:
Relation between frequency and length:
The frequency of the fundamental mode of vibration of a wire is inversely proportional to its length when the tension and mass per unit length are constant. i.e. f ∝ 1/l
Relation between frequency and tension:
The frequency of the fundamental mode of vibration of a wire is directly proportional to the square root of its tension when the length and mass per unit length are constant. i.e. f ∝ √T
Relation between frequency and mass per unit length:
The frequency of the fundamental mode of vibration of a wire is inversely proportional to the square root of its mass per unit length when the length and tension are constant. i.e. f ∝ 1/√m
Using the above relations:
For the initial length of the wire:
f1 ∝ 1/l1
For the shortened length of the wire:
f2 ∝ 1/l2
Since the tension is constant, we can equate the two relations:
f1/f2 = l2/l1
Substituting the given values, we get:
200/f2 = 500/1000
f2 = 400Hz
Therefore, the fundamental frequency of transverse vibration of the wire when shortened to 500mm at the same tension is 400Hz.
Answer:
Option d) 400Hz.
The frequency of the fundamental mode of transverse vibration of stret...
Frequency (n) =1/2L under root T/m so from This N1L1=N2L2 so 200*1000=N2*500 so from this N2=400Hz
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