Physics 112 Quiz 1 Ch 16

When a person inhales helium gas, the speed of sound waves through the helium gas is greater compared to air. The wavelength of sound depends on the speed of sound, so with the increased speed in helium, the wavelength of the person's speech will also increase. Since frequency is inversely proportional to wavelength, the increase in wavelength will result in a decrease in frequency. Therefore, the person's speech will have lower frequencies, making it sound deeper or lower than their normal speech in air.

Ultrasound refers to sound waves of high frequencies. High frequency sound waves have shorter wavelengths and are capable of penetrating through different materials. Ultrasound is commonly used in medical imaging to create images of internal organs and tissues. It is also used in various industrial applications such as cleaning, measuring distance, and detecting flaws in materials. Therefore, the correct answer is high frequencies.

When the sound enters an environment where the speed of sound is greater, the wavelength (lambda) becomes longer. This is because the speed of sound is directly proportional to the wavelength. As the speed of sound increases, the wavelength also increases to maintain a constant frequency (f). This is known as the wavelength-frequency relationship, where wavelength is inversely proportional to frequency. Therefore, if the speed of sound increases, the wavelength must also increase in order to keep the frequency constant.

As the singer practices the vocal scale singing "do, re, mi, fa, so, la, ti, do", the sounds produced will have progressively shorter wavelengths and higher frequencies. This is because as the pitch increases, the frequency of the sound waves also increases, resulting in shorter wavelengths.

As you drink soda from the bottle, the amount of soda decreases, leaving more empty space inside the bottle. Blowing across the top of the bottle creates a sound by causing the air inside to vibrate. When there is more soda in the bottle, the air column above it is shorter, resulting in a higher frequency and therefore a higher pitch sound. As you remove more soda, the air column becomes longer, causing a decrease in frequency and a lower pitch sound. Therefore, the lower the pitch of the resulting sound corresponds to the more soda you have removed.

When you hold a seashell to your ear, you hear the "sound of the sea" because sounds whose intensities are too small to be audible under normal conditions are enhanced by reflection within the shell, producing noise similar to the ocean. The shape and structure of the shell act as a natural amplifier, reflecting and amplifying the sound waves that enter the shell. This phenomenon creates the illusion of hearing the sound of the sea when in reality it is just the amplified ambient noise around us.

In order to have resonance and a standing wave on the string of a violin or guitar, there should be nodes at both ends. Nodes are points of zero displacement in a standing wave, and they occur at the ends of the string. This is because the ends of the string are fixed and cannot vibrate, resulting in zero displacement at those points. Having nodes at both ends allows for the formation of a standing wave pattern, which is necessary for resonance to occur.

The sound intensity of 1.0E-2 W/m at a distance of 100 m can be used to calculate the power output of the explosion. The surface area of a sphere is 4πR^2, so the power output can be found by multiplying the sound intensity by the surface area at that distance. By plugging in the values, we get 1.0E-2 W/m * 4π(100^2) = 1.3E3 W. Therefore, the total power output of the explosion is 1.3E3 W.

The resonant frequency of a tube closed at one end can be calculated using the formula f = (n * v) / (4 * L), where f is the frequency, n is the harmonic number, v is the speed of sound, and L is the length of the tube. In this case, the length of the tube is given as 0.027 m and the speed of sound is given as 344 m/s. Since the tube is closed at one end, the harmonic number is 1. Plugging these values into the formula, we get f = (1 * 344) / (4 * 0.027) = 3200 Hz, which is equivalent to 3.2E3 Hz. Therefore, the resonant frequency of the tube is 3.2E3 Hz.