The fundamental frequency of a closed organ pipe of length `20 cm` is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends isA. 80 cmB. 100 ccmC. 120 cmD. 140 cm

The fundamental frequency of a closed organ pipe of length `20 cm` is

1.	The fundamental frequency of a closed organ pipe of length `20 cm` is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends isA. 80 cmB. 100 ccmC. 120 cmD. 140 cm
Answer» Correct Answer - c The fundamental frequencies of closed organ pipe are given as, `v_(c)=v/(4l)` `because` Second overtone of an open organ pipe `v_(2)=(3v)/(2l_(1))` Where `l_(1)=` length of the open pipe `because v_(c)=v_(2)` `rArr v/(4l)=(3v)/(2l_(1))rArr l_(1)=6l=6xx20` `therefore l_(1) = 120` cm

Discussion

No Comment Found

Related InterviewSolutions

In a sonometer wire, the tension is maintained by suspending a 50.7 kg mass from the free end of the wire. The suspended mass has a volume of 0.0075 m 3. The fundamental frequency of the wire is 260 Hz . If the suspended mass is completely submerged in water, the fundamental frequency will become (take `g = 10 ms^(-2)`) [A. 220 HzB. 230 HzC. 240 HzD. 260 Hz
Fundamental frequency of pipe is 100 Hz and other two frequencies are 300 Hz and 500 Hz thenA. Pipe is open at both the endsB. Pipe is closed at both the endsC. One end open and another end is closedD. None of the above
A stretched wire of lenth 110 cm is divided into three segments whose frequencies are in ratio `1 : 2 : 3.` Their length must beA. 20 cm, 30 cm, 60 cm,B. 60 cm, 30 cm, 20 cmC. 60 cm, 20 cm, 30 cm,D. 30 cm, 60 cm, 20 cm
The integral multiple of fundamental frequencies . Which are actully present in the sound isA. HarmonicsB. ResonanceC. OvertonesD. Forced vibration
When a string is divided into three segments of lengths `l_(1),l_(2) and l_(3)` the fundamental frequencies of these three segments are `v_(1), v_(2) and v_(3)` respectively. The original fundamental frequency (v) of the string isA. `sqrt(v)= sqrt(v_(1))+sqrt(v_(2))+sqrt(v_(3))`B. `v= v_(1) + v_(2)+v_(3)`C. `1/v=1/v_(1) + 1/v_(2)+1/v_(3)`D. `1/sqrt(v)= 1/sqrt(v_(1))+1/sqrt(v_(2))+1/sqrt(v_(3))`
When the string of the sonometer of length l between the bridges vibrates in the first overtone, then the antinode formed from one end of the string will beA. `lamda/2`B. `lamda/4 " and " (3lamda)/(4)`C. `lamda/6, (3lamda)/(6) " and " (5lamda)/(6)`D. `lamda/8, (3lamda)/(8) " and " (7lamda)/(8)`
The length of the stretched sonometer wire is 1m, if the fundamental frequency are in the ratio 1:2 . Where should be the bridge be placed to divide the wire two segments ?A. `1/3 : 2/3`B. `2/3:1/3`C. `3/2:3`D. `2:3`
A sonometer wire of length 100cm has fundamental frequency of 320Hz. Then the velocity of the transverse waves is ,A. 640 m/sB. 160 m/sC. 320 m/sD. 480 m/s
A sonometer wire , `100 cm` in length has fundamental frequency of `330 Hz`. The velocity of propagation of tranverse waves along the wire isA. 330 m/sB. 660 m/sC. 990 m/sD. 115 m/s
One metre long sonometer wire is stretched with a force of 4kg wt, another wire of same material and diamter is arranged along a side. The second wire is stretched with a force of 16kg wt. if the length of the second wire is in its second harmonic is the same as fifth harmonic of the first wire, then the length of the second wire will beA. 40cmB. 40cmC. 80cmD. 70cm

The fundamental frequency of a closed organ pipe of length `20 cm` is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends isA. 80 cmB. 100 ccmC. 120 cmD. 140 cm

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment