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Compare this problem with the preceding problem. A plane sound wave in air at \(20^{\circ} \mathrm{C},\) with wavelength \(589 \mathrm{mm},\) is incident on a smooth surface of water at \(25^{\circ} \mathrm{C},\) at an angle of incidence of \(3.50^{\circ} .\) Determine the angle of refraction for the sound wave and the wavelength of the sound in water..

Short Answer

Expert verified
The angle of refraction for the sound wave is 14.1° and the wavelength of the sound in water is 2.57 m.

Step by step solution

01

Calculate the speed of sound in air at 20°C

Use the equation for the speed of sound in air: \( v_{air} = (331.4 + 0.6T) m/s \), where \( T \) is the temperature in Celsius. So, \( v_{air} = 331.4 + 0.6 \times 20 = 343 \, m/s \).
02

Find the frequency of the wave in air

The equation that relates wavelength, wave speed, and frequency is: \( v = f \cdot \lambda \). Solve this equation for \( f \) to find the frequency: \( f = v / \lambda \). So, \( f = 343 / 0.589 = 582 \, Hz \). Note that frequency is a property of the source of the wave and doesn't change when the wave goes from one medium into another.
03

Calculate the speed of sound in water at 25°C

Use the empirical equation for the speed of sound in water: \( v_{water} = 1402.7 + 4.7T - 0.043T^2 \, m/s \), where \( T \) is the temperature in Celsius. So, \( v_{water} = 1402.7 + 4.7 \times 25 - 0.043 \times 25^2 = 1496 \, m/s \).
04

Calculate the angle of refraction using Snell's Law

Snell's Law states that \( sin(i) / sin(r) = v_{air} / v_{water} \). Solve it for \( r \) to find the angle of refraction: \( r = arcsin[(v_{water} / v_{air}) \times sin(i)] \). So, \( r = arcsin[(1496 / 343) \times sin(3.5)] = 14.1° \).
05

Calculate the wavelength of sound in water

Knowing the frequency and the speed of sound in water, we can calculate the wavelength in water using the equation \( v = f \cdot \lambda \). Solve for \( \lambda \) to find the wavelength: \( \lambda = v / f \). So, \( \lambda = 1496 / 582 = 2.57 \, m \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Sound in Different Mediums
The speed of sound varies significantly based on the medium it travels through because sound is a mechanical wave that requires a material to propagate. In general, sound travels fastest in solids, slower in liquids, and slowest in gases. This is due to the density and elasticity of the medium influencing how quickly particles can transfer the sound energy.

Temperature also affects the speed of sound, as seen with air at different temperatures in the original problem. The speed of sound in air increases with temperature because warmer air has more energy, causing particles to vibrate faster and transmit sound quicker. A formula to estimate the speed of sound in air at different temperatures is:
\[ v_{air} = (331.4 + 0.6 \times T) \text{ m/s} \]
where \( T \) is the temperature in Celsius. Liquids, such as water, have a different relationship with temperature, and an empirical equation is often used to compute the speed of sound in water:
\[ v_{water} = 1402.7 + 4.7 \times T - 0.043 \times T^2 \text{ m/s} \]
Understanding how sound speed changes with mediums and temperatures helps predict how sound waves will behave in various environments.
Snell's Law for Sound
Snell's Law is a principle that describes how waves, including sound waves, bend when they pass from one medium to another. This bending, known as refraction, occurs due to the change in the wave's speed as it enters the new medium.

For sound, Snell's Law is represented as:\[ \frac{\sin(i)}{\sin(r)} = \frac{v_{medium 1}}{v_{medium 2}} \]
where \( i \) is the angle of incidence, \( r \) is the angle of refraction, and \( v_{medium 1} \) and \( v_{medium 2} \) are the speeds of sound in the two respective mediums. You can use Snell's Law to predict the direction of a sound wave when it crosses into a different medium. As seen in the exercise, the speed of sound in air and water allowed us to calculate how much the sound wave will bend or change direction.
Frequency of a Sound Wave
Sound waves are characterized by their frequency, which is the number of waves that pass a point in a given period of time, usually measured in Hertz (Hz). The frequency determines the pitch of a sound—the higher the frequency, the higher the pitch.

One crucial aspect to understand is that the frequency of a sound wave is determined by the source of the wave and remains constant when the wave travels from one medium to another. This constancy was highlighted in the exercise when calculating the frequency of the sound wave in air with:
\[ f = \frac{v}{\lambda} \]
where \( f \) is frequency, \( v \) is the speed of the sound, and \( \lambda \) is the wavelength. Regardless of changes in speed and wavelength when entering a new medium, the frequency stays the same.
Wavelength of a Sound Wave
The wavelength of a sound wave is the distance between consecutive crests or troughs in the wave. It's inversely related to frequency—sound waves with higher frequencies have shorter wavelengths.

In our exercise, the wavelength changes when the sound wave enters water due to the difference in the speed of sound. The frequency remains constant, but because sound travels faster in water than air, the wavelength increases. We used the speed of sound in water and the constant frequency to find the new wavelength with the formula:\[ \lambda = \frac{v}{f} \]
Hence, the wavelength of the sound wave in water was calculated and showed a significant increase from its original value in air. This concept is essential for understanding how sound behaves differently in various mediums, affecting how we perceive and measure sound.

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Most popular questions from this chapter

Builders use a leveling instrument with the beam from a fixed helium-neon laser reflecting in a horizontal plane from a small flat mirror mounted on an accurately vertical rotating shaft. The light is sufficiently bright and the rotation rate is sufficiently high that the reflected light appears as a horizontal line wherever it falls on a wall. (a) Assume the mirror is at the center of a circular grain elevator of radius \(R .\) The mirror spins with constant angular velocity \(\omega_{m} .\) Find the speed of the spot of laser light on the wall. (b) What If? Assume the spinning mirror is at a perpendicular distance \(d\) from point \(O\) on a flat vertical wall. When the spot of laser light on the wall is at distance \(x\) from point \(O,\) what is its speed?

The distance of a light bulb from a large plane mirror is twice the distance of a person from the plane mirror. Light from the bulb reaches the person by two paths. It travels to the mirror at an angle of incidence \(\theta,\) and reflects from the mirror to the person. It also travels directly to the person without reflecting off the mirror. The total distance traveled by the light in the first case is twice the distance traveled by the light in the second case. Find the value of the angle \(\theta\).

(A).Consider a horizontal interface between air above and glass of index 1.55 below. Draw a light ray incident from the air at angle of incidence \(30.0^{\circ} .\) Determine the angles of the reflected and refracted rays and show them on the diagram. (b) What If? Suppose instead that the light ray is incident from the glass at angle of incidence \(30.0^{\circ} .\) Determine the angles of the reflected and refracted rays and show all three rays on a new diagram. (c) For rays incident from the air onto the air-glass surface, determine and tabulate the angles of reflection and refraction for all the angles of incidence at \(10.0^{\circ}\) intervals from \(0^{\circ}\) to \(90.0^{\circ} .\) (d) Do the same for light rays coming up to the interface through the glass.

The shoreline of a lake runs east and west. A swimmer gets into trouble \(20.0 \mathrm{m}\) out from shore and \(26.0 \mathrm{m}\) to the east of a lifeguard, whose station is \(16.0 \mathrm{m}\) in from the shoreline. The lifeguard takes negligible time to accelerate. He can run at \(7.00 \mathrm{m} / \mathrm{s}\) and swim at \(1.40 \mathrm{m} / \mathrm{s} .\) To reach the swimmer as quickly as possible, in what direction should the lifeguard start running? You will need to solve a transcendental equation numerically.

The Apollo 11 astronauts set up a panel of efficient corner cube retro- reflectors on the Moon's surface. The speed of light can be found by measuring the time interval required for a laser beam to travel from Earth, reflect from the panel, and return to Earth. If this interval is measured to be \(2.51 \mathrm{s}\) what is the measured speed of light? Take the center-to-center distance from Earth to Moon to be \(3.84 \times 10^{8} \mathrm{m},\) and do not ignore the sizes of the Earth and Moon.

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