Ncert Exercises & Solutions

15.17 A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m/sec).

It is given in the question that :

Length of the pipe, l = 20 cm = 0.2 m

Frequency of the source = n^{th} normal mode of frequency,

ν_{n} = 430 Hz

Speed of sound, v = 340 m / s

In a closed pipe, the n^{th} normal mode of frequency :

\begin{array}{l} ν_{n} = (2 n - 1) \frac{v}{4 l}; n is an integer = 0, 1, 2, 3 . . . . . \\ 430 = (2 n - 1) \frac{340}{4 \times 0 .2} \\ 2 n - 1 = \frac{430 \times 4 \times 0 .2}{340} = 1 .01 \\ 2 n = 2 .01 \\ n \approx 1 \end{array}

Hence, the first mode of vibration frequency is resonantly excited by the given source .

In a pipe open at both ends :

n = \frac{2 {lν}_{n}}{v} = \frac{2 \times 0 .2 \times 430}{340} = 0 .5

Since the number of the mode of vibration (n) has to be an integer,

the given source does not produce a resonant vibration in an open pipe .

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions