AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

For a closed organ pipe, the frequency of fundamental mode is vi=‌

v

4Lc

where, v is the velocity of sound in air and Lc is the length of the closed organ pipe.
For an open organ pipe, the frequency of fundamental mode is
Ve=‌

V

2Le

where, L0 is the length of the open pipe.
∵‌‌‌‌Lc=Lo(given)
∴‌‌v0=2vc...(i)
vo−vc=2...(ii)
Solving Eqs. (i) and (ii), we get
ve=4Hz and vc=2Hz
Now, length of open pipe is halved, then its frequency of fundamental mode is
v0=‌

v

2(‌ Le 2 )

=2(‌

v

2L0

)
=2vv=2×4=8Hz
When the length of the closed pipe is doubled, its frequency of fundamental mode is
vi′=‌

v

4(2Lo)

=‌

1

2

(‌

v

4Lv

)
=‌

1

2

×2=1Hz
Hence, number of beats produced per second =v0−vc′=8−1=7Hz