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A wire under tension vibrates with a frequency of 450 cycles per sound. What would be the fundamental frequency if the wire were half as long, twice as thick and under one fourth tension (a) 225 cps (b) 190 cps (c) 247 cps (d) 174 cps

A wire under tension vibrates with a frequency of 450 cycles per sound. What would be the fundamental frequency if the wire were half as long, twice as thick and under one fourth tension

(a) 225 cps

(b) 190 cps

(c) 247 cps

(d) 174 cps

Grade:12

2 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
11 months ago
576-2161_a.PNG
Vikas TU
14149 Points
11 months ago
As per the question,
Given data:
A wire under tension vibrates with a frequency = 450 hertz
According to the speed of the string, we have
v = sqrt[(T-l)/m]
Let V₁ = 450 Hz
Now,
The wire were half as long twice as thick and under one-fourth tention:
New length = l/4
m = 2m
V1 = sqrt[(T-l)/m]

And
V2 = sqrt[(T-l)/2m-4]

Divide V₂ by V₁, we get

∴ v₂ = 159.12 Hz
Hence, the fundamental frequency = 159.12 Hz

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