Question

A tuning fork of frequency 340 Hz is vibrated just above the tube of 120 cm height. Water is poured slowly in the tube. What is the minimum height of water necessary for the resonance? ( speed of sound in air = 340 m/s )
A. $$45\mathrm{c}\mathrm{m}$$
B. $$30\mathrm{c}\mathrm{m}$$
C. $$40\mathrm{c}\mathrm{m}$$
D. $$25\mathrm{c}\mathrm{m}$$

Answer 1

Hint:The concept of resonance in the sound waves from the tuning fork The mathematical relation for the wavelength as well as the frequency is utilised to resolve the problem.

Complete Step by Step Answer:The concept of resonance in the sound waves from the tuning fork The mathematical relation for the wavelength as well as the frequency is utilised to resolve the problem.
Given:
The frequency of the tuning fork is, $$f=340\mathrm{H}\mathrm{z}$$.
The height of the tube is, $$h=120\mathrm{c}\mathrm{m}$$.
The speed of sound in air is, $$v_s=340\mathrm{m}/\mathrm{s}$$.
 The wavelength of the sound waves from the tuning fork is,
$$\begin{array}{l}\lambda ={\displaystyle \frac{v}{f}}\\ \lambda ={\displaystyle \frac{340\mathrm{m}/\mathrm{s}}{340\mathrm{H}\mathrm{z}}}\\ \lambda =1\mathrm{m}\end{array}$$
 The first resonating length is,
$$L_1={\displaystyle \frac{\lambda }{4}}$$
Substituting the value of $$\lambda$$ in the above equation as,
 $$\begin{array}{l}{L}_1={\displaystyle \frac{\lambda }{4}}\\ L_1={\displaystyle \frac{1}{4}}\\ L_1=0.25\mathrm{m}\\ L_1=0.25\mathrm{m}=0.25\mathrm{m}\times {\displaystyle \frac{100\mathrm{c}\mathrm{m}}{1\mathrm{m}}}=25\mathrm{c}\mathrm{m}\end{array}$$
The second resonating length is,
$$L_2={\displaystyle \frac{3\lambda }{4}}$$
Substituting the value of $$\lambda$$ in the above equation as,
 $$\begin{array}{l}{L}_2={\displaystyle \frac{\lambda }{4}}\\ L_2={\displaystyle \frac{3\times 1\mathrm{m}}{4}}\\ L_2=0.75\mathrm{m}=0.75\mathrm{m}\times {\displaystyle \frac{100\mathrm{c}\mathrm{m}}{1\mathrm{m}}}=75\mathrm{c}\mathrm{m}\end{array}$$
The third resonating length is,
$$L_3={\displaystyle \frac{5\lambda }{4}}$$
Substituting the value of $$\lambda$$ in the above equation as,
 $$\begin{array}{l}{L}_3={\displaystyle \frac{5\lambda }{4}}\\ L_3={\displaystyle \frac{5\times 1\mathrm{m}}{4}}\\ L_3=0.125\mathrm{m}=0.125\mathrm{m}\times {\displaystyle \frac{100\mathrm{c}\mathrm{m}}{1\mathrm{m}}}=125\mathrm{c}\mathrm{m}\end{array}$$
The above result clearly shows that the third resonance is not possible because it is exceeding the length of the tube of 120 cm.
The minimum height of water necessary for the resonance is,
$$\begin{array}{l}{h}_m=120\mathrm{c}\mathrm{m}-L_2\\ h_m=120\mathrm{c}\mathrm{m}-75\mathrm{c}\mathrm{m}\\ h_m=45\mathrm{c}\mathrm{m}\end{array}$$
Therefore, the minimum height of water necessary for the resonance is 45 cm and option A is correct.

Note:The mathematical relation and the fundamentals of resonance of sound through tuning fork is to be remembered. Along with fundamental concepts the key relation of consecutive resonating length is also remembered.