Q141 / 160 · AP EAMCET 2024 (18 May Shift 1) · Physics

Two stretched strings A and B when vibrated together produce 4 beats per second. If the tension applied to the string A increased, the number of beats produced per second is increased to 7. If the frequency of string B is 480 Hz initially. The frequency of string A is

  • A
    473 Hz
  • B
    476 Hz
  • Correct answer
    C
    484 Hz
  • D
    487 Hz

Solution

$\mathrm{n}_{\mathrm{A}} \sim \mathrm{n}_{\mathrm{B}}=4$ Now, if tension in string A increased, then frequency of string A increased and beats produced increased. $\begin{array}{ll}\therefore & \mathrm{n}_{\mathrm{A}}\gt\mathrm{n}_{\mathrm{B}} \\ \therefore & \mathrm{n}_{\mathrm{B}}=480 \mathrm{H}_2 \\ \therefore & \mathrm{n}_{\mathrm{A}}=\mathrm{n}_{\mathrm{B}}+4=480+4=484 \mathrm{~Hz}\end{array}$

Hence, the correct answer is option: 'C'.

Relevant Questions