Question

An organ pipe closed at one end has a length of 1 meter and an organ pipe has a length$1.6\,m$. Speed of sound in air is $320\,m/s$. The two pipes can resonate for a sound of frequency
(A)100 Hz
(B) 240 Hz
(C) 320 Hz
(D) 400 Hz

Answer 1

In this solution, we will calculate the gravitational force exerted by the ring on the sphere. The force exerted by one on the other will have the same magnitude but opposite directions.

Formula used: In this solution, we will use the following formula:
-Resonant frequencies for closed pipe: $f = \dfrac{{nv}}{{4l}}$ where $n$ has odd values
- Resonant frequencies for open pipe: $f = \dfrac{{nv}}{{2l}}$ where $n$ are all positive integers

Complete step by step answer:
In the situation given to us, we want to find a frequency for which both the pipes can resonate. To find such a frequency, let us find the first few frequencies of resonance for both the pipes.
For the closed pipe, the frequencies of resonance can be found from
$f = \dfrac{{nv}}{{4l}}$
Substituting $v = 320\,m/s$ and $l = 1\,m$, we get
$f = n(80)$ where $n$ are odd.
So the first few frequencies are
$f = 80,240,400....$
For the open pipe, the frequencies of resonance will be
$f = \dfrac{{nv}}{{2l}}$ where $n$ are all integers
Substituting $v = 320\,m/s$ and $l = 1.6\,m$, we get
$f = n(100)$
So the first few frequencies are
$f = 100,200,300,400\,Hz$
So we can see that both the open and the closed pipe will resonate at 400 Hz. For the open pipe, the order or resonance will be the fifth harmonic and for the open pipe, it will be the fourth harmonic.

Hence the correct choice is option (D).

Note: While both the pipes resonate for 400 Hz, they can also resonate for higher frequency values but they are not given in the options. Also, the frequency at which both pipes will resonate doesn't need to correspond to the same order of resonance for both the pipes as well.