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Elan Coil [88]
4 years ago
5

What is the fundamental frequency of a particular medium ?

Physics
2 answers:
VLD [36.1K]4 years ago
5 0
From my research, the question has the following choices:

a.the lowest frequency at which a standing wave is possible
b. the highest frequency at which s standing wave is possiblec. the only frequency at which a standing wave is possible
d. the only frequency at which standing wave is not possible

From there, the correct answer is A.
sladkih [1.3K]4 years ago
5 0

Answer:

This is the minimum possible frequency generated by the given medium

Explanation:

Let say the medium through which sound is produced is a hollow pipe closed at one end

Now at minimum possible situation the wave that will form in it will be of quarter length so we can say

L = \frac{\lambda}{4}

so we have

\lambda = 4L

now we will have

f = \frac{v}{\lambda}

so fundamental frequency of closed pipe will be

f = \frac{v}{4L}

similarly if pipe is closed then in that case the wavelength will be half inside the pipe

L = \frac{\lambda}{2}

now we have

f = \frac{v}{2L}

in this way we can say it is the minimum frequency of sound produced by the source

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Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

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