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Question

Physics Question on Waves

An organ pipe AA, with both ends open, has fundamental frequency 300Hz300\, Hz. The third harmonic of another organ pipe BB, with one end open, has the same frequency as the second harmonic of pipe AA. The lengths of pipe AA and BB are (speed of sound in air = 343m/s)343\, m/s)

A

57.2 cm and 42.9 cm

B

57. 2 cm and 45.8 cm

C

42.9 cm and 32.2 cm

D

42.9 cm and 34.0 cm

Answer

57.2 cm and 42.9 cm

Explanation

Solution

Given that fundamental frequency of open organ pipe (A)=300Hz(A)=300\, H z
f1=300Hz=v2l1f_{1}=300 \,H z=\frac{v}{2 l_{1}} \ldots (i)
Frequency of second harmonic-of open organ pipe (A)=vl1(A)=\frac{v}{l_{1}}
and frequency of third harmonic of closed organ pipe =3v4l2=\frac{3 v}{4 l_{2}}
Now according to question vl1=3v4l2\frac{v}{l_{1}}=\frac{3 v}{4 l_{2}}
l1l2=(43)\Rightarrow \frac{l_{1}}{l_{2}}=\left(\frac{4}{3}\right) \ldots (ii)
Form E (i), we get
300=3432l1300=\frac{343}{2 l_{1}}
l1=343600=52.16=57.2cml_{1}=\frac{343}{600}=52.16=57.2 \,cm
Now from E (ii), we get
l2=34×l1l_{2}=\frac{3}{4} \times l_{1}
=34×57.2=42.9cm=\frac{3}{4} \times 57.2=42.9 \,cm