AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

For a closed organ pipe, the frequency of fundamental mode is $v_i = \frac{v}{4 L_c}$ where, $v$ is the velocity of sound in air and $L_c$ is the length of the closed organ pipe. For an open organ pipe, the frequency of fundamental mode is $V_e = \frac{V}{2 L_e}$ where, $L_0$ is the length of the open pipe. $\because \;\; L_c = L_o$(given) $\therefore \;\; v_0 = 2 v_c$...(i) $v_o - v_c = 2$...(ii) Solving Eqs. (i) and (ii), we get $v_e = 4 \text{ Hz}$ and $v_c = 2 \text{ Hz}$ Now, length of open pipe is halved, then its frequency of fundamental mode is $v_0 = \frac{v}{2 \left( \frac{L_e}{2} \right)} = 2 \left( \frac{v}{2 L_0} \right)$ $= 2 v_v = 2 \times 4 = 8 \text{ Hz}$ When the length of the closed pipe is doubled, its frequency of fundamental mode is $v_i' = \frac{v}{4 (2 L_o)} = \frac{1}{2} \left( \frac{v}{4 L_v} \right)$$= \frac{1}{2} \times 2 = 1 \text{ Hz}$ Hence, number of beats produced per second $= v_0 - v_c' = 8 - 1 = 7 \text{ Hz}$