Physics test 2 (ch 15,ch 16)

An organ pipe is made to play a low note at 27.5 HzHz, the same as the lowest note on a piano. A) Assuming a sound speed of 343 m/sm/s, what length open-open pipe is needed? B) What length open-closed pipe would suffice?

λ = v / f = 343 m/s / 27.5 s = 12.5 m A) open ended, L = λ/2, soL = 12.5 m / 2 = 6.24 m B) closed ended, L = λ / 4, soL = 12.5 m / 4 = 3.12 m

A train whistle is heard at 320 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 310 Hz. A) What is the speed of the train before slowing down? B) What is the speed of the train after slowing down?

(f+1)/(f+2), solve for Vs

A large solar panel on a spacecraft in Earth orbit produces 1.9 kW of power when the panel is turned toward the sun. A) What power would the solar cell produce if the spacecraft were in orbit around Saturn, 9.5 times as far from the sun?

21

A child has an ear canal that is 1.3 cmcm long. A) At what sound frequencies in the audible range will the child have increased hearing sensitivity?

A

An experimenter finds that standing waves on a string fixed at both ends occur at 12 HzHz and 16 HzHz , but at no frequencies in between. A) What is the fundamental frequency? B) Select the standing-wave pattern for the string at 16 HzHz .

A B f0 x m=16 f0 = 4 so m = 4 or 4 loops

A 14 kgkg hanging sculpture is suspended by a 95-cmcm-long, 4.0 gg steel wire. When the wind blows hard, the wire hums at its fundamental frequency. What is the frequency of the hum? A) What is the frequency of the hum?

A)

A vuvuzela is a simple horn, typically 0.65 mm long, that fans use to make noise at sporting events(Figure 1). A) What is the frequency of the fundamental note produced by a vuvuzela? Assume a sound speed of 343 m/sm/s.

A)

An earthquake 20 km from a city produces P and S waves that travel outwards at 5000 and 3000 m/s, respectively. A) Once city residents feel the shaking of P wave, how much time do they have before the S wave arrives?

A)

In 2003, an earthquake in Japan generated 1.1 Hz waves that traveled outward at 7.0 km/s. 200 km to the west, seismic instruments recorded a maximum acceleration of 0.25 gg along the east-west axis. A) How much time elapsed between the earthquake and the first detection of the waves? b) Was this a transverse or a longitudinal wave?

A)

A 70 cm -long steel string with a linear density of 1.3 g/mg/m is under 200 NN tension. It is plucked and vibrates at its fundamental frequency. A) What is the wavelength of the sound wave that reaches your ear in a 20∘C room?

A) linear density m = 1.1 g/m = 1.1 *10 ^-3 kg / m Tension T = 200 length L = 90 cm = 0.9 m frequency f = ( 1/ 2L ) √[ T / m ] substitute values we get f = 236.890 Hz wavelength of the sound = speed of sound /frequency = ( 343 m / s ) / ( 236.890 Hz ) = 1.448 m

The first formant of your vocal system can be modeled as the resonance of an open-closed tube, the closed end being your vocal cords and the open end your lips. A) Estimate the frequency of the first formant from the graph in (Figure 1). b)Estimate the length of the tube of which this is a resonance.

A) B)

A spider spins a web with silk threads of density 1300 kg/m3kg/m3 and diameter 3.0 μmμm . A typical tension in the radial threads of such a web is 7.0 mNmN. Suppose a fly hits this web. A) Which will reach the spider first: the very slight sound of the impact or the disturbance traveling along the radial thread of the web? b) What is the speed of disturbance traveling along the radial thread of the web?

A) Disturbance traveling along the thread. B)

When in air, a waterproof speaker emits sound waves that have a frequency of 1000 Hz A) When the speaker is lowered into water, does the frequency of the sound increase, decrease, or remain the same? When the speaker is lowered into water, does the frequency of the sound increase, decrease, or remain the same? b) When the speaker is lowered into water, does the wavelength of the sound increase, decrease, or remain the same?

A) Frequency is property of source and it doesn't change with medium. So, frequency will be constant here. B) When speaker is lowered into water, velocity of sound increases. Since, velocity = wavelength * frequency wavelength will also increase

Using a dish-shaped mirror, a solar cooker concentrates the sun's energy onto a pot for cooking. A cooker with a 2.0-mm-diameter dish focuses the sun's energy onto a pot with a diameter of 25 cm A) Given that the intensity of sunlight is about 1000 W/m2W/m2, how much solar power does the dish capture? B) What is the intensity at the base of the pot?

A) Power = Intensity x area I = 1000 W/m2 1000x (pi(1)/2)^2)) This is how much solar power the dish captures. B) Part B. Intensity is given by: Intensity = Power/Area Power captured by dish = 3100 W Surface area of pot = pi*r^2 = pi*d^2/4 d = diameter of pot = 25 cm Area = pi*0.25^2/4 = 0.049 m^2 Now Intensity will be I = 3100/0.049 I = 64114 W/m^2

A loudspeaker emits a sound wave at a particular frequency. A) If the intensity of the wave is doubled, how does the speed of the wave change? B) If the frequency of the wave is doubled, how does the speed of the wave change?

A) Speed of the wave doesn't depend on intensity of wave, or frequency of wave. Speed of wave depends only on the properties of the medium through which it is propagating. Therefore speed of wave will remain unaffected in both cases

Two identical loudspeakers 2.0 mm apart are emitting 1800 HzHz sound waves into a room where the speed of sound is 340 m/sm/s . A) Is the point 4.0 mm directly in front of one of the speakers, perpendicular to the plane of the speakers, a point of maximum constructive interference, perfect destructive interference, or something in between?

A) The other speaker is a distance of d^2 = 2^2 + 4^2 d = sqrt(20) = 2sqrt(5) m Path difference from the two speakers to the point is 2sqrt(5) - 4 = 4.47 - 4 = 0.47 m If this path difference is a whole number of wavelengths we get constructive interference and if it is a whole number of half wavelengths we get descructive interference. wavelength = v/f = 340/1800 = 0.188 m 0.47/0.188 = 2.5 0.47/0.094 = 5 so it is destructive interference.

The displacement of a wave traveling in the positive x-direction is y(x,t) = (3.5cm)cos(2.7x−92t), where x is in m and t is in s. A) What is the frequency of this wave?

A) angular frequency (w) = 2pif 92 rad = 2pif f 92 rad/2pi = f

The frequency of light emitted from hydrogen present in the Andromeda galaxy has been found to be 0.10%% higher than that from hydrogen measured on earth. A) Is this galaxy approaching or receding from the earth? B) At what speed is this galaxy approaching the earth?

A) approaching earth B) pic

A) An FM radio station broadcasts at a frequency of 101.3 MHzMHz. What is the wavelength? b) What is the frequency of a sound source that produces the same wavelength in 20∘∘ CC air?

A) c = 3 × 10^8 m/s is the speed of light in vacuum, B) speed of sound in air at 20 C is 343 m/s

Two identical loudspeakers 2.10 mm apart are emitting sound waves into a room where the speed of sound is 340 m/sm/s. Abby is standing 3.50 mm in front of one of the speakers, perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. A) What is the lowest possible frequency of sound for which this is possible?

A) d = n (wL) d = n (v/f) f = n (v/d) lowest, = (v/d)

You are standing 2.50 mm directly in front of one of the two loudspeakers shown in the figure. They are 3.00 mm apart and both are playing a 686 HzHz tone in phase. A) As you begin to walk directly away from the speaker, at what distances from the speaker do you hear a minimum sound intensity? The room temperature is 20 ∘C∘C.

A) find wavelength distribute after adding X over and squaring!!

The displacement of a wave traveling in the positive x-direction is y(x,t) = (3.5cm)cos(2.7x−92t), where x is in m and t is in s. B) What is the wavelength of this wave?

B) k = 2pi/WL WL = 2pi/k k = wave number = 2.7 m (-1) solve for WL

A sun-like star is barely visible to naked-eye observers on earth when it is a distance of 7.0 light years, or 6.6×10^16m, away. The sun emits a power of 3.8×10^26W. A) Using this information, at what distance would a candle that emits a power of 0.20 W just be visible?

Given 1. Distance between the star and the observer - 6.6×10^16 m 2. The sun like star emits a power of - 3.8×10^26W 3. power emitted by candle - 0.20 W

The African cicada is the world's loudest insect, producing a sound intensity level of 107 dB at a distance of 0.50 m. A) What is the intensity of its sound (in W/m2W/m2) as heard by someone standing 2.6 mm away?

Intensity = 1/(distance)^2

A laser beam has intensity I0. A) What is the intensity, in terms of I0, if a lens focuses the laser beam to 1/10 its initial diameter? B) What is the intensity, in terms of I0I0, if a lens defocuses the laser beam to 10 times its initial diameter?

Intensity in this case is the beam power (P) per unit of area (A) .. I = P/A (W/m²) For the same emitted power, the intensity is inversely proportional to the area As area A ∝ d² .. • d/10 reduces the area by a factor 100 .. ►intensity increases by a factor of 100 • 10d increases the area by a factor 100 .. ►intensity decreases by a factor of 100 -Part A: I = 100, Part B: I = .01.

The figure shows a standing wave oscillating at 100 HzHz on a string a) What is the wave speed?

L = (3/2)λ =>λ=2L/3 = 40cm Wave speed = v = λf = 40*100 = 4000cm/sec = 40m/sec

log rules

One important rule for manipulating logarithms is the product rule: log(xy)=log(x)+log(y) Consider a simple example to illustrate this rule. You know from the previous discussion that log(102)=2log⁡(102)=2. Since 102=10×10102=10×10, you could rewrite the original logarithm using the product rule: log(102)=log(10)+log(10)=1+1=2log⁡(102)=log⁡(10)+log⁡(10)=1+1=2.

An echocardiogram uses 4.4 MHz ultrasound to measure blood flow in the aorta. The blood is moving away from the probe at 1.2 m/sm/s . A) What is the frequency shift of the reflected ultrasound? Use values given in the table if necessary. Enter a positive value if the received frequency is higher than the emitted frequency, and a negative value if the received frequency is lower than the emitted frequency.

Reflection from a moving ob when Vo << v 1540

Elephants can communicate over distances as far as 6 km by using very low-frequency sound waves. A) What is the wavelength of a 10 Hz sound wave emitted by an elephant? Assume air temperature is 20∘C∘C.

Speed of sound in air V = 343 m/s Wavelength corresponding to frequency given Lambda = v / frequency Lambda = 343/10 Lambda = 34.3 m

The first dancer in the line is 10 m from the speaker playing the music; the last dancer in the line is 120 m from the speaker. Approximately how much time elapses between when the sound reaches the nearest dancer and when it reaches the farthest dancer?

Speed of sound in air=340m/s Assume At a certain time the first dancer heard the sound at t=0 After that sound has to travel (120-10) m distance to the farthest dancer We know that ,for uniformly propagating sound wave, time taken=Distance travelled/velocity of sound Therefore ,t=110/340=0.323 approximately equal to 1/3 seconds

A dark blue cylindrical bottle is 22 cm high and has a diameter of 7.0 cm . It is filled with water. The bottle absorbs 60%% of the light that shines on it as it lies on its side in the noonday sun, with intensity 1000 W/m2 A) By how much will the temperature of the water increase in 5.0 minmin if there's negligible heat loss to the surrounding air? The specific heat of water is 4190 J/(kg⋅K)

Steps 1. area receiving light wxh 2. energy/sec = Intensity x area 3. only 60 percent actually 4. for 5 minutes x5x60 5. V of cylinder pir^2h in cm^3 ---> mass in g 6 solve for delta t mcdeltaT Answer is in kelvin not celsius

A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length LL is 40.5 cmcm , 54.0 cmcm , and 67.5 cmcm A) What is the frequency of the tuning fork? The air temperature is 20C

The difference between each of the lengths is ∆L = 16.7 cm = 0.167 m ∆L = 75.9 cm − 59.2 cm = 16.7 cm For an open-open tube, the fundamental wavelength is λ = 2∆L. λ = 2 x 16.7 cm = 2 x 0.167 m = 0.334 m The frequency is f = v/λ f = 343 m/s / 0.334 m f = 1027 Hz f = 1.027 KHz

A) What is the value of log(1,000,000)log⁡(1,000,000)? B) If a speaker gives a sound intensity of 10^−6 W/m2 at a certain point, what is the sound intensity level βbeta at that point? c) The power in the speaker from Part A is doubled, bringing the sound intensity to 2×10−6W/m2. This leads to β=10log(2×106). Which of the following equations is equivalent to log(2×106)?

The logarithm of x , written log(x) , tells you the power to which you would raise 10 to get x . So, if y=log(x) , then x=10y . It is easy to take the logarithm of a number such as 102 , because you can directly see what power 10 is raised to. That is, log(102)=2 . a) 6 b) Intensity I = 10^-6 W/m^2 Sound intensity level β = 10 log[I/Io] Here Io = 10^-12 W/m^2 β = 10 log[10^-6 W/m^2/10^-12 W/m^2] = 10 * log[10^6] = 10 * 6 = 60 c) log(2)+log(106)

shows history graphs of two different points on a string as a wave pulse moves along the string. The blue curve is the history graph for the point at xx = 1.0 cmcm, and the green curve is for the point at xx = -3.0 cmcm. a) What is the velocity (including the correct sign for its direction) of this wave?

We know that Velocity = delta x /delta t = [-3 -1]cm/[5-0]s = - 0.8 cm/s answer

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a coyote is facing a bird and listening to it whistle at 1150 HzHz . The bird is 2.6 mm away, directly in front of the coyote's right ear. The coyote's ears are 15 cmcm apart on a line perpendicular to the right ear-bird line. a) What is the difference in the arrival time of the sound at the left ear and the right ear? Hint: You are looking for the difference between two numbers that are nearly the same. What does this near equality imply about the necessary precision during intermediate stages of the calculation? b) What is the ratio of this time difference to the period of the sound wave?

a) b)

LASIK eye surgery uses pulses of laser light to shave off tissue from the cornea, reshaping it. A typical LASIK laser emits a 1.0-mmmm-diameter laser beam with a wavelength of 193 nm. Each laser pulse lasts 11 nsns and contains 1.8 mJ of light energy a) What is the power of one laser pulse? b) During the very brief time of the pulse, what is the intensity of the light wave?

a) B)

Parasaurolophus in (Figure 1) was a dinosaur whose distinguishing feature was a hollow crest on the head. The 1.5-mm-long hollow tube in the crest had connections to the nose and throat, leading some investigators to hypothesize that the tube was a resonant chamber for vocalization. a) If you model the tube as an open-closed system, what are the first three resonant frequencies? Assume a speed of sound of 350 m/s

a) For a half open pipe fn = n*v/4L (n = 1,3,5,.....) For a half open pipe the resonant frequencies are 1 ,3 ,5... so f1 = 350/(4*1.7) = 51.47Hz f3 = 3*51.47 = 154.41Hz f5 = 5*51.47 = 257.35Hz

To study the physical basis of underwater hearing in frogs, scientists used a vertical tube filled with water to a depth of 1.9 mm. A microphone at the bottom of the tube was used to create standing sound waves in the water column. Frogs were lowered to different depths where the standing waves created large or small pressure variations. Because the microphone creates the sound, the bottom of the tube is a pressure antinode; the water's surface, fixed at atmospheric pressure, is a node. a) What is the fundamental frequency of this water-filled tube? b) A frog sits on a platform located 0.38 mm from the bottom. What is the lowest frequency that would result in a sound node at this point?

a) V/4L!! b) v =1480

A concert loudspeaker suspended high off the ground emits 39 WW of sound power. A small microphone with a 1.0 cm2cm2 area is 60 mm from the speaker. A) What is the sound intensity at the position of the microphone? B) What is the sound intensity level at the position of the microphone?

power, P=39 W area, A=1cm^2 distance, r=60m A) intensity, I=P/A =P/(4pi*r^2) =39/(4pi*60^2) =8.62*10^-4 w/m^2 B) sound intensity level=10*log(I/Io) =10*log(8.62*10^-4/(1*10^-12)) =89 dB

The wave speed on a string under tension is 250 m/s. A) What is the speed if the tension is doubled?

sqrt(2) x 250 = 354

People with very good pitch discrimination can very quickly determine what note they are listening to. The note on the musical scale called C6C6 (two octaves above middle CC) has a frequency of 1050 Hz. Some trained musicians can identify this note after hearing only 12 cycles of the wave. A) How much time does this correspond to?

t = (number of cycles)/frequency (cycles/second)

A sun-like star is barely visible to naked-eye observers on earth when it is a distance of 7.0 light years, or 6.6×10^16m, away. The sun emits a power of 3.8×10^26W. A) Using this information, at what distance would a candle that emits a power of 0.20 W just be visible?

understand

The displacement of a wave traveling in the positive x-direction is y(x,t) = (3.5cm)cos(2.7x−92t), where x is in m and t is in s. C) What is the speed of this wave?

v = fxWL 15 x 2.3

Bats use echolocation to navigate. They can emit ultrasonic waves with frequencies as high as 1.0×105 Hz A) What is the wavelength of such a wave? The speed of sound in air is 340 m/s

velocity of wave = f*lambda v = 340 f = 100000hz wavelength = 340/100000 = 3.4*10^-3 m