Test Index

Waves

Section: Physics

Question : 12 of 31

Marks: +1, -0

Solution:

let

n_1

harmonic is corresponding to 50.7 cm &

n_2

harmonic is corresponding

83.9

\therefore \frac{\lambda}{2} = (83.9-50.7)\text{cm}\;\;\;\; \frac{\lambda}{2} = 33.2\text{cm}

\lambda = 66.4\text{cm}\;\;\;\; \therefore \frac{\lambda}{4} = 16.6\text{cm}

Length corresponding to fundamental mode must be close to

\frac{\lambda}{4}

50.7

cm must be an odd multiple of this length

16.6 \times 3 = 49.8

cm.therefore

50.7

is 3rd harmonic

If end correction is e, then

e + 50.7 = \frac{3\lambda}{4}\text{cm}

Speed of sound,

v = f\lambda

\therefore v = 500 \times 66.4

cm/sec

= 332.000

m/s

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