Question

The frequency of the fundamental note of a 66 cm
long open tube is the same as that of a 20 cm long
stretched wire whose mass is 0.1 g cm . If the speed
of sound in air is 330 m s-, then determine the
tension in the wire.
Ans. 100 N.​

Answer 1

Answer:

Tension in the wire is 100 N

Explanation:

Frequency of the wave in stretched wire is given as

$f = \frac{1}{2L}\sqrt{\frac{T}{m/L}}$

also for the frequency in open tube

$f = \frac{v}{2L}$

here we know

$v = 330 m/s$

$L = 66 cm$

so we have

$f = \frac{330}{2 \times 0.66}$

$f = 250 Hz$

now from the formula of string

$250 = \frac{1}{2(0.20)}\sqrt{\frac{T}{0.01}}$

$\frac{T}{0.01} = 10000$

$T = 100 N$

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Topic : Waves

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