Two sound waves with wavelengths 5.0 m and 5.5. m respectively, each propagate in a gas with velocity 330 m/s. We expect the following number of beats per second :-

Question

A steel wire with mass per unit length \( 7.0 \times 10^{-3} \, \text{kg/m} \) is under a tension of \( 70 \, \text{N} \). The speed of transverse waves in the wire will be:

Answer 1

We are given two sound waves with wavelengths of 5.0 meters and 5.5 meters, respectively. Both waves are propagating in a gas with a velocity (speed of sound) of 330 meters per second. We need to determine the number of beats per second.

The formula for calculating the frequency f of a wave given the velocity v and wavelength \lambda is:

f = \frac{v}{\lambda}

Now, we will calculate the frequencies of both sound waves:

For the first wave with wavelength 5.0 m:
f_1 = \frac{330}{5.0} = 66 \text{ Hz}
For the second wave with wavelength 5.5 m:
f_2 = \frac{330}{5.5} \approx 60 \text{ Hz}

Beats per second, also known as the beat frequency, is given by the absolute difference between the two frequencies:

\text{Number of Beats per second} = |f_1 - f_2|

Substituting the frequencies we calculated:

\text{Number of Beats per second} = |66 - 60| = 6 \text{ Hz}

This means we expect 6 beats per second.

Conclusion: The correct answer is 6.

Answer 2