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Question: What is the base frequency if a pipe gives notes of frequency \(425,255\) and \(595\) and decides wh...

What is the base frequency if a pipe gives notes of frequency 425,255425,255 and 595595 and decides whether it is closed at one end or open at both ends?
A. 17,closed17,closed
B 85,closed85,closed
C. 17,open17,open
D. 85,open85,open

Explanation

Solution

We need to know the full concept of close pipe and open pipe then the only way you can solve this problem easily. First assuming it as a close pipe using the close pipe concept you will come to know the base frequency now with the help of that base frequency we can generate the relation with the given frequency.

Complete step by step answer:
Open pipe: It is a pipe that is open at both the ends and a standing wave can form if the wavelength of the sound allows there to be an antinode.
Close pipe: But in this one end is open and the other is closed similarly here also standing waves with sound of an appropriate frequency.

As per the given problem, three notes of frequency are given =425,255 = 425,255 and595595. Now let us assume that the base frequency is nn that base frequency is for a closed pipe. As we know the concept of close pipe is that notes of close pipe are n,3n,5nn,3n,5n etc. As in the problem three notes of frequency is given applying these frequency in close pipe concept we get,
255=3n255 = 3n
⇒n=2553\Rightarrow n = \dfrac{{255}}{3}
⇒n=85\Rightarrow n = 85
Hence we get the base frequency as 8585 .

Now with the help of this base frequency we can compare the other two given frequencies.
Here,
5n=4255n = 425
This is because,
5×85=4255 \times 85 = 425
Similarly,
7n=5957n = 595
This is because,
7×85=5957 \times 85 = 595
Hence from the above discussion we conclude that the base frequency is for a close pipe with base frequency 8585 .

Therefore the correct option is (B)\left( B \right).

Note: Generally this type of question is logical, we have to analyse the given frequency and have to assume the pipe and the second important in this is proper comparison of given frequency with the assumed pipe. If your assumption is wrong then the alternate one will be correct and once cross check it with another pipe if your assumption is wrong before choosing the option.