SOLUTION SET 5

1. A wave has crests and troughs. From the top of one crest to the bottom of the next trough is 13 cm. The length from crest to trough along the x axis is 28 cm. Find the a. amplitude and b. wavelength.
Solution:
a. Since the top of one crest to the bottom of the next trough is 13 cm, and since the distance from the center (zero) line to the apex (crest, top, etc.) is the Amplitude, and since the center line is halfway from the apex to the trough, we multiply the 13 cm by ½, or, ½ (13 cm) = A = 6.5 cm.
b. Since the length from crest to trough along the x-axis is 28 cm, and since the definition of wavelength the distance from crest to crest, and the distance from crest to trough is only half a wavelength, then we have to double the distance given to find the wavelength, l = 2 x 28 = 56 cm.

2. The speed of surface waves in water decreases as the water becomes more shallow. Imagine water waves crossing the surface of an otherwise calm lake at a speed of v1 = 2.0 m/s, with a wavelength of l1 = 1.5 meters. When these same waves approach shore, the speed decreases to v2 = 1.6 m/s, while the frequency remains the same. Find the wavelength, l2, in the shallow waters.
Solution: We have learned that by multiplying the wavelength, l, by the frequency, n, we get that “famous” equation, resulting in the velocity: l n = v. To find the non-changing frequency, we can rewrite the equation as n = v1/l1.= (2.0)/(1.5) = 1.33 Hz. Remember, n 1 = n 2 = n = 1.33 Hz. So, using the same equation that l n = v., to find, , we just divide both sides by thfrquency, n. And the get this: l 2 = v2 /n.= (1.6 m/s)/(1.33) = 1.2 meters.

3. Suppose that you wanted to double the wave speed, from v to 2v, on a tight string. The force that is keeping it tight is called “tension” but it is still a force and has units of Newtons. How much would you have to increase the tension so the velocity would double?
Solution: We remember from the laboratory exercise, “Tin Can Phone with Soundwaves on a String,” that the velocity is: v = (T/m)½ , where T is the tension or force, and m is the linear density. So, if we wanted the velocity to go from v to 2v, then we would have to quadruple the tension (i.e., four times more, because: 2V = (4T/d)½

4. Waves on a particular string travel at v1 = 16 m/s. To double that speed to v2 = 32 m/s, how much would you have to increase the tension?
Solution:
The clever student should quickly realize that this is the same identical problem as Number 9, so the answer is, again, . quadruple the tension.

5. Write an expression for a harmonic wave, such that amplitude, A = 0.16 m, wavelength l = 2.1 m, and period of P = 1.8 seconds.
Solution: On page in the book, it says that the expression for a harmonic wave is:
y = A cos(2px/l – 2pt/P), and thus, if you plug in the data given, your answer becomes:
y(x,t) = 0.16 Cos[(2px/2.1) – (2pt/1.8)]

6. A soundwave in air has a frequency of n1 = 425 Hz.
a. Find its wavelength
b. If its frequency is increased, does the wavelength increase, decrease, or remain unchanged
c. Calculate its wavelength if the frequency is n2 = 475 Hz.
Solution: Once again, we use the “famous” equation, l n = v., and to find wavelength, we divide both sides by the frequency. To get l = v/n.

a. l = v/n. = (342 m/s)/(425 Hz) = 0.805 meters. (but where did we get the speed, v? This is a sound wave. In air. The speed of sound in air is always 342 m/s.
b.l n = vsound
c. l = 342/475 = 0.72 meters.

7. A man throws a rock down to the bottom of a well. From the instant that it leaves his hand until the moment he hears the rock hit bottom, a period of 1.20 seconds passes. The depth of the well is 8.85 meters. Find the initial speed, vi, of the rock (the instant that it leaves the man's hand).
Solution: We are looking for vi, the initial speed of the rock (the instant that it leaves the man's throwing hand). So, what do we know? We know that the final speed of the rock, vf = 0.0 m/s (it is stopped when it hits the well's bottom). Maybe we can use the formula:

vf2 - vi2 = 2 a y, where since vf = 0.0, we can re-write it as:

- vi2 = 2 a y. We know the number “2.” We know the distance down, y = 8.85 meters, but do we know the acceleration, a?

What is “a” ? a = g + a’. And g is Earth's gravitational pull, while a' is the extra acceleration added by the man's hand. Since I don't know where this is going, I am going to put this path to solving it on “hold” and try something else.

We know from a previous lab that the average speed as it falls is related to the initial speed, i.e., vave = ½ vi

We need to find the average velocity. We know that speed is distance over time, or, v = y/t. Let's call:

t1 = the time it takes for the object to reach the bottom after being thrown
t2 = the time for the object’s sound to reach the top after making noise at the bottom.

We do know that total time from the time the rock was released until the sound reached the man's ears is given: t1 + t2 = 1.2 s.

We know that the speed of sound here is: vs = y/t2 ; so, t2 = y/vs = 8.85 m/342 m/s = 0.026 s. If t2 = 0.26 seconds, then t1 can be found by subtraction.

t1 + 0.026 = 1.2 s
t1 = 1.2 s - 0.026 = 1.174 s

Now we can find the average speed, as the rock was falling/thrown down: distance/time, or y/t1, or vave = 8.85/1.174 = 7.54 m/s

and, if vave = ½ vi, then 2vave = vi..= (2)(7.54) = 15.08 m/s. Rounding, we get:

vi = 15 m/s.

8. Twenty identical violins are being played by 20 identical violinists at the same identical volume, for an aggregate sound level of 82.5 dB.
a. Find the dB level of one of the violins.
b. Find the dB level of twice as many identical violins (40).
Solution: According to the table provided in the book (or on the test), one sees that a dB level of 80 has an energy flux of 10-4 Watts/m2.
a. If 20 violins has an energy flux of 10-4 Watts/m2, then one violin would have 1/20th of that, or, 10-4 /20 = 100 x 10-6 / 20 = 5 x 10-6. Rechecking the table shows that an energy flux of that amount would be the equivalent to about 66 dB.
b. If 20 violins has an energy flux of 10-4 Watts/m2, then 40 violins would have twice that, or, 2 x 10-4 Watts/m2, and checking the chart, that energy flux would give that a reading of about 85 dB.

9. Find the weight (in Newtons) of the air in your physics class. Assume that the room is a cube of 4 meters on a side, and that the density of air is rair = 1.29 kg/m3.
Solution:
Volume 4 x 4 x 4 = 64 m3 ; rair = 1.29 kg/m 3. The density, r, is defined to be:
r = mass/volume.

So, the mass = r vol = (1.29)(64) = 82.56 kg. Finally, the “weight” in Newtons is its force in gravity: F = m g = (82.56)(9.8) = 809 Newtons.

10. You buy a gold ring at a pawn shop, and run a “test” to determine if it's pure gold. The ring's mass is m = 0.014 grams, and has a volume of 0.022 cm3. The density of gold is rAu = 19.3 kg/m3.
Solution: Density is mass divided by volume, and it this case, r = M/V = (0.014g/0.0022 cm3). = 6.6 g/cm3. However, since rAu = 19.3 kg/m3. You can be sure that the ring is NOT solid gold.

11. Crutches will often have rubber tips at the bottom, so that contact with a flooring surface will decrease scratching of the floor, increase friction, and its broader tip spreads the pressure out over a larger area. Assume the radius of the tip of the cane to be 1.2 cm, and that the radius of the rubber tip is 2.5 cm. What is the ratio of the pressure of the cane alone, P1, to the pressure with the rubber tip, P2: P1/P2.
Solution:
Pressure = Force/Area
Area p r2
Area 1 = p r2
Area 2 = p (2.5/1.2)2r2 =p (2.08)2r2
Thus, Area 2 is 4.34 greater than Area 1, and thus, the pressure in Area 2 is 1/(4.34) as great; in other words, is 0.23 the pressure. P1/P2. = 4.34.

12. Two drinking glasses, G1 and G2, are filled with water to the same level of 5.0 cm. The diameter of the base of G1 is twice the diameter of the base of G2, but otherwise the two glasses are identical.
a. Is the weight of the water in G1 greater, less, or the same as weight in G2?
b. Is the water pressure at the bottom of G1 greater, less, or the same as the water pressure at the bottom of G2?
Solution: There is more water in G1 than G2, but the same area
a. G1 has a greater weight.
b. Water pressure is identical.

13. On Wednesday, August 15, 1934, William Beebe and Otis Barton made history by descending in the Bathysphere, a steel sphere 4.75 feet in diameter, to 3028 feet below sea level.
a. as the Bathysphere was lowered, was the buoyant force exerted on it at a depth of 10 feet greater, less, or equal to it at 50 feet?
b. Choose the best explanation from below.
I Magic
II Miracle
III Mayonnaise.
Solution:
a. Same
b. III

14. A fish called Wanda is carrying a pebble in its mouth swims with a small constant velocity in a small bowl. When the fish drops the pebble to the bottom of the bowl, does the water level rise, fall or stay the same?
Solution: Stays the same.

15. Explain what a a hydrometer is.
Solution:
How would I know? I'm not an auto mechanic.

A hydrometer is an instrument used to measure the relative density of liquids to water, which has a density of r = 103 kg/m3. It comes from two words, hydro, meaning “water,” and metros meaning “to measure.”
Hydrometers are typically made of glass, which, itself is a liquid. The device consists of a cylindrical stem and a bulb weighted with the element mercury (not the planet or the car or the Roman god). This allows the stem and bulb to float upright. The liquid to be tested (such as honey or antifreeze) is poured into a tall jar, and the hydrometer is gently lowered into the liquid – the honey or antifreeze, etc., until it floats freely. The point at which the surface of the liquid touches the stem of the hydrometer is noted. Hydrometers usually contain a paper scale inside the stem, so that the relative density can be read directly. The scales may be of different types, as hydrometer units have not yet been standardized by the SIU.
Hydrometers may be calibrated for different uses, such as a lactometer for measuring the density (creaminess) of milk, a saccharometer for measuring the density of sugar in a liquid, or an alcoholometer for measuring higher levels of alcohol in liquor and other distilled spirits.

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