Question

At 16°C, two open-end organ pipes, when sounded together produce 34 beats in 2 seconds. How many beats per second will be produced if the temperature rises to 51°C? Neglect the increase in the length of the pipes.

• (A) 18
• (B) 15
• (C) 10
• (D) 5

Answer 1

The Correct Option is A

Explanation:
Given:Number of beats produced per second at $16°C=\frac{34}{2}=17$We have to find the number of beats per second that will be produced if the temperature rises to 51°CLet$L_1,L_2=$ lengths of the two organ pipes$v_1,v_2=$ lowest (or fundamental) frequencies emitted by the pipes at 16°C$v_1^{\prime },v_2^{\prime }$ = lowest frequencies emitted by the pipes at 51°C$v_{16},v_{51}=$ velocities of sound at 16°C and 51°C respectively.The expression for fundamental frequency for a tube open at both ends is given by$v=\frac{v}{2L}$$v_1=\frac{v_{16}}{2L_1}$ and $v_2=\frac{v_{16}}{2L_2}$Beat frequency at ${16}^{\circ }\mathrm{C}=v_1-v_2=17$ (given)$\Rightarrow \frac{v_{16}}{2L_1}-\frac{v_{16}}{2L_2}=17$$\Rightarrow v_{16}[\frac{1}{2L_1}-\frac{1}{2L_2}]=17$....(i)Let the number of beats per second at ${51}^{\circ }\mathrm{C}=\mathrm{b}$Then, $v_1-v_2=b$$\Rightarrow \frac{v_{51}}{2L_1}-\frac{v_{51}}{2L_2}=b$$\Rightarrow v_{51}[\frac{1}{2L_1}-\frac{1}{2L_2}]=b$....(ii)Dividing eq. (ii) by eq. (i), we get$\frac{v_{51}}{v_{16}}=\frac{b}{17}$$\sqrt{\frac{51+273}{16+273}}=\frac{b}{17}[\because v_t\propto \sqrt{t+273}]$$\Rightarrow b=17\times \sqrt{\frac{324}{289}}=17\times \frac{18}{17}$$=18{\mathrm{s}}^{-1}$Hence, the correct option is (A).