Question 1

The speed of sound in water is 4.3 times the speed of sound in air. A whistle produces a sound wave in air with a frequency f₀. When this sound wave enters the water, its frequency will be

Accepted Answer

The frequency of a sound wave does not change when it moves from one medium to another; only its speed and wavelength change.

Question 2

Two piano strings are simultaneously played-their mass and length are identical, but the tensions differ. If the first string resonates with a fundamental frequency of 330 Hz, and its tension is 1.2% greater than that of the second string, what is the fundamental frequency of the second string?

Accepted Answer

The fundamental frequency of a string is proportional to the square root of the tension. If the tension in the first string is 1.2% greater, then the frequency of the second string will be lower. The relationship between frequency and tension is given by $f \propto \sqrt{T}$, so if the tension in the first string is 1.012 times the tension in the second, the frequency ratio is $\sqrt{1.012}$. Thus, the frequency of the second string is $330 / \sqrt{1.012}$ ≈ 328 Hz.

Question 3

A string with a mass per length of 2.00 g/m is stretched between two points that are 0.400 m apart. The frequency of the second mode of the stretched string is 50.0 Hz lower than the frequency of the third mode. What is the tension in the string?

Accepted Answer

The answer of A string with a mass per length...

Question 4

A string with a mass per length of 2.00 g/m is stretched between two points that are 0.400 m apart. The frequency of the third mode of the stretched string is 120 Hz higher than the frequency of the fundamental. What is the tension in the string?

Accepted Answer

The frequency of the fundamental mode $f_1$ can be found using the following formula for the frequency of a stretched string: 
$$f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$
where $L$ is the length of the string, $T$ is the tension, and $\mu$ is the mass per length. Plugging in the given values, we get: 
$$f_1 = \frac{1}{2(0.400	ext{ m})}\sqrt{\frac{T}{2.00	ext{ g/m}}} = \frac{1}{0.800	ext{ m}}\sqrt{\frac{T}{0.00200	ext{ kg/m}}}$$
Solving for $T$ gives: 
$$T = \frac{4f_1^2\mu L^2}{\pi^2} = \frac{4(f_1+120	ext{ Hz})^2\mu L^2}{\pi^2}$$
Substituting the given values and solving for $T$ gives us the answer: 
$$T = \boxed{	extbf{(C) }4.61	ext{ N}}$$

Question 5

A sound wave radiates from a source uniformly in all directions. If the power of the sound source is 200 watts, then the intensity of the sound wave 100 m from the source is

Accepted Answer

The answer of A sound wave radiates from a source...

Question 6

A string with a mass per length of 2.00 g/m is stretched between two points that are 0.400 m apart. The frequency of the third mode of the stretched string is 20.0 Hz lower than the frequency of the fourth mode. What is the tension in the string?

Accepted Answer