Question

hey ppl plz ans this...

Two instruments having stretched strings are being played in union. When the tension of one of the instruments is increased by 1%, 3 beats are produced in 2s. the initial frequency of vibration of each wire is
300 Hz b) 500 Hz c) 1000 Hz d) 400 Hz

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Answer 1

ans is 300hz .. because

Frequency of vibration of syring

isf=V2l=12l×Tμ−−√=cT−−√∆ff×100=12×∆TT×100∆ff×100=12×1=0.5∆ff=0.5×10−2∆f=0.5×10−2×fbeat frequency is =0.5×10−2×fbut beat frequency is=320.5×10−2×f=32f=320.5×10−2=300HzRegards..

hope it will help.

Answer 2

Answer :-

✨Provided :-

▪️Increase in Tension of string = 1%

▪️Beats produced = 3 beats in 2 sec

◾Now we have

$frequency = \dfrac{1}{2(length)} \sqrt{\dfrac{Tension}{\dfrac{mass}{length}}}$

Or

$f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\dfrac{m}{l}}}$

◾Now let the initial frequency be "x"

▪️Then

$x = \dfrac{1}{2l}\sqrt{\dfrac{T}{\dfrac{m}{l}}}$

$x = \dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}$

◾And the frequency after increase in frequency be " x' "

▪️Then

→$x' = \dfrac{1}{2l}\sqrt{\dfrac{T + 1\% T}{\dfrac{m}{l}}}$

→$x' = \dfrac{1}{2l}\sqrt{\dfrac{\dfrac{101}{100} T}{\dfrac{m}{l}}}$

→$x' = \dfrac{1}{2l}\sqrt{\dfrac{\dfrac{101}{100}Tl}{m}}$

→$x' = \dfrac{1}{2l}\sqrt{\dfrac{101Tl}{100m}}$

→$x' = \sqrt{\dfrac{101}{100}} \times \dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}$

▪️As $x = \dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}$

→$x' = \dfrac{10.05}{10} \times x$

→$x' = 1.005 \times x$

→ $x' = x + (0.005)x$....(i)

▪️Also

$x' = x + \dfrac{3}{2}$ ......(ii)

▪️By taking (i) and (ii)

$\implies x + (0.005)x = x + \dfrac{3}{2}$

$\implies x + (0.005)x - x = \dfrac{3}{2}$

$\implies (0.005)x = \dfrac{3}{2}$

$\implies \dfrac{5}{1000}x = \dfrac{3}{2}$

$\implies x = \dfrac{3}{2}\times \dfrac{1000}{5}$

$\implies x = \dfrac{3000}{10}$

$\implies x = 300$

⏩So initial frequency = 300hz