22.19 To determine the wavelength, use the fact that the speed of a wave is equal to its wavelength times its frequency

Transcription

1 hhh.schaums.22.19_ To determine the wavelength, use the fact that the speed of a wave is equal to its wavelength times its frequency or speed = waveln gth frequency speed is in m/s, wavelength is in meters, and frequency is in hertz where one hertz is equal to an inverse second (frequency is a measure of how many oscillations occur in one second) Same as (Notice that the speed of the radio waves is the same as the speed of light. Radio waves are a form of electromagnetic radiation just as light is. Radio waves have a lower frequency than light) First convert the wavelength from cm to meters. Then use speed = waveln gth frequency Convert the wavelength to meters. The frequency is 120 Hz. Multiply to get the speed. (b) The speed of a wave on a string is equal to speed = please memorize this formula

2 Linear density is equal to the mass of the string divided by its length (hence the term linear density) Using the determined speed from part a and the that is given in the problem, solve for the linear density using speed = Once you get the linear density you can solve for the mass of the string using its length (given) and the fact that LD = mass(kg) length(m) Notice that you must first convert the length to meters and then finally convert the mass to grams amplitude - Distance of vertical displacement from zero point to its highest point frequency - is the inverse of the period of the wave. The period of the wave is how long it takes the wave to go through one full wavelength. In this problem we are told that the wave is generated by a 60 cycle per second (60 hz) generator. Wavelength Length of one full wave speed derive form the formula speed = frequency wavelength period inverse of frequency First you must find the mass of the wire. To do this you must find its volume first. The wire is a cylinder and the volume of a cylinder is equal to its (cross sectional area pi r^2) times its

3 length. Once you have found its volume in cm 3 you must convert this to m 3 3 cm 3 1m = 100cm notice that you must cube the conversion factor. Thus there are 1,000,000 cm 3 in one m 3. Now that you have the volume in m 3, you can use the density of the copper and its volume to find the mass of the copper. Density (conventional) = mass volume Now you have the mass of the wire in kg. Divide this by the length of the wire in meters to get the linear density. To find the in wire, recall that a two kilogram mass is hung from it. The in the wire is equal to the weight (mg) of this mass. Now you have the linear density and the, you can use speed = to solve for the speed of the wave Since the wave vibrates in three segments, the wavelength of the wave is equal to 2/3 of the length of the string, or 120 cm. The frequency of the wave is 270 hz. Use this information to determine the speed of the wave.

4 22.26 The speed remains constant. Initial condition speed = frequency 2 / 3length = 165hz (0.667) Final condition speed = frequency 1/ 2length = x 0.50 Dividing the two equations we get speed = speed x 0.5 the left side of the equation is simply 1, solve for x Determine the mass of the cable by dividing its weight by 9.8 m/s 2 Determine the linear density of the cable Determine the speed of the wave (convert kn to newtons). The wave must travel 2 times the length of cable (up and back). Since distance = velocity(speed) x time, determine the time it takes to travel 60 m Since speed = and frequency is proportional to speed find the new frequency by using

5 frequency new = frequency old ( 1 4) 4 mass 1 length 2 = frequency old 2 16 This gives a different answer than the one in the book, which I believe is incorrect. If you make the wire twice as thick, then its mass will be four times its initial mass (since its radius will double and area is proportional to the square of the radius. Thus volume will increase by a factor of 4 as will the mass of the wire) The frequency is inversely proportional to the square of the linear density. Since the density of steel is lower than silver the frequency of steel s fundamental will be higher than silver s by a ration equal to the inverse of the square root of the ratio of the density of steel to silver (try chewing on that one a while). Now the book gives the volume density and not the linear density but these ratios wills still be the same since the diameter is the same The first overtone has two times the frequency of the fundamental. So the fundamental has a frequency of 100 hz. The wavelength of the fundamental is 1.2 m and the linear density is kg/ 0.6 m = 5.0 x 10-3 T kg/m. Then just use v = LD (a) think of a guitar string. The fundamental sounds when you pluck it in the middle (which is an antinode) (b) The first three overtones are shown in figure 22-2 on page 222 (a lot of 2 s huh). To make the string resonate the loudest you always pluck it where there is a node. BY THE WAY, IN PHYSICS THE FUNDAMENTAL IS CALL THE FIRST HARMONIC AND THE THE FIRST OVERTONE IS CALLED THE SECOND HARMONIC (it depends on which physics book you use as to whether the term harmonic or overtone is used Just another way to keep you confused If the rod is clamped at its center, each end is an anti-node and the center is a node. THUS THE RESONATING BAR IS EQUAL TO 1/2 A WAVELENGTH. The wavelength is just the speed divided by the frequency, and thus the bar must be 1/2 of this length (see example 22.1 on page 226 and figure 22-3 on page 227) See figure 22-4 on page 227. The end of the bar must be an antinode so the first overtone corresponds to each side of the bar being 3/4 of a wavelength. So the entire bar is 1 1/2 wavelengths

6 antinode putting clamp here would give the same results antinode Since the first overtone is 1200 hz, the bar is clamped as in the first figure above. The bar is equal to 1.5 wavelengths as can be seen from the figure above, Thus the wavelength is 4.0 m. The speed of the wave is equal to the wavelength times the frequency The jar is like a pipe that is open on one end. One end of the jar is a node of a wave and the other end is an anti-node. Thus the jar (or pipe closed at one end) is 1/4th of the wave length

7 (A pipe that is open on both ends has an anti-node on both ends and thus is 1/2 the wavelength of the soundwave produced) OK, so the wavelength is m. From speed = wavelength times frequency you can get the speed of sound Now for the second part the closed end of the pipe is a node and now the open end is the second antinode of the wave. The distance from a node of a wave to the second anti node is 3/4 of a wavelength, so the pipe will resonate at 3/4 of a wavelength as well as at 1/4 of a wavelength.