Question

A guitar string under the tension of 200N, vibrating in its fundamental mode, gives 6 beats/sec. With a tuning fork the player increase the string tension to 242N and again gets 6 beat/sec. Find the frequency of the tuning fork?

Answer 1

Answer on Question #72214-Physics-Other

A guitar string under the tension of 200N, vibrating in its fundamental mode, gives 6 beats/sec. With a tuning fork the player increases the string tension to 242N and again gets 6 beat/sec. Find the frequency of the tuning fork?

Solution

$$f_2 - f = 6 \frac{\text{beat}}{s}.$$ $$f - f_1 = 6 \frac{\text{beat}}{s}.$$ $$\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}$$ $$\sqrt{\frac{T_2}{T_1}} f_1 - f = 6 \frac{\text{beat}}{s}.$$ $$\sqrt{\frac{T_2}{T_1}} (f - 6) - f = 6.$$ $$\left(\sqrt{\frac{T_2}{T_1}} - 1\right) f = 6 \left(\sqrt{\frac{T_2}{T_1}} + 1\right)$$ $$f = 6 \frac{\left(\sqrt{\frac{T_2}{T_1}} + 1\right)}{\left(\sqrt{\frac{T_2}{T_1}} - 1\right)} = 6 \frac{\left(\sqrt{\frac{242}{200}} + 1\right)}{\left(\sqrt{\frac{242}{200}} - 1\right)} = 126 \text{ Hz}.$$

Answer: 126 Hz.