Chapter 15, example problems:


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1 Chapter, example problems: (.0) Ultrasound imaging. (Frequenc > 0,000 Hz) v = 00 m/s. λ <.0 mm. Frequenc required > 00 m/s /.0 mm = Hz. (Smaller wave length implies larger frequenc, since their product, being equal to the sound velocit, can not change. (.) Speed of propagation vs. particle speed. Eq. (.): (x, t) = A cos[ω (x/v t)] = A cos [πf (x/v t)]. (a) Putting in ω = v k, k = π / λ, and using the fact that v (x/v t) = (x v t), we obtain: (x, t) = A cos[( π / λ) (x v t)]. (b) The transverse velocit of a particle in the string on which the wave travels: The transverse displacement of a particle at position x on that string as a function of t is (x, t) = A cos[( π / λ) (x v t)], so the transverse velocit of that particle as a function of t is: v (x, t) = {A cos[( π / λ) (x v t)] } / t x = A sin[( π / λ) (x v t)] ( π / λ) ( v) = ( π f) Α sin[( π / λ) (x v t)]. (c) The maximum transverse speed of a particle in the string, v max, is just the amplitude of the above velocit wave, namel π f Α (= ω Α). (This is the same expression as that for the maximum velocit of a harmonic oscillator. This is not a coincidence. Ever point in a wave is indeed executing a simple harmonic motion, onl with a phase which varies with x.) This maximum transverse velocit of a particle in the string is in general different from the wave velocit. For them to be equal, we would have to require ω Α = v, or, since ω = v k, we would have to require kα =, which just means Α = λ / π. For this maximum transverse velocit to be less or greater than the wave velocit, wee need A to be less or greater than λ / π. (In realit, A is practicall alwas much less than λ / π. Thus the maximum transverse velocit of a particle in the string is practicall alwas much less than the wave velocit.) (.6).0 m string, weight. N. Tied to the ceiling. Lower end supports a weight W. Plucked slightl. Wave traveling up the string obes: (x, t) = (8.0 mm) cos(7 m x 70 s t)] (thus x is measured upward). (a) Time for a pulse to travel the full length of the string: We need the wave velocit, which is 70 s / 7 m =.87 m/s. Then the travel time is.0 m /.87 m/s = 0.09 s. (b) Actuall, this problem is correct onl if the weight of the string,. N, is much less than the weight W, and therefore can be neglected with respect to the latter. Then the tension in the string, F, is due to W alone, and is therefore equal to W. The linear mass densit of the string, μ, is (. N / 9.8 m/s ) /.0 m = kg/m. Then using the formula for the velocit of a transverse wave on a string, v = (F / μ), we can find W = v μ = (.87 m/s) kg/m =. N, which is indeed much larger than. N, confirming the validit of the approximation we made in the beginning. (But the error is about %.)
2 (c) The number of wavelengths on the string at an instant of time =.0 m / (π / 7 m ) =.. (d) Equation for waves traveling down the string is: (x, t) = A cos [( π / λ) (x m/s t)]. (That is, A and λ can be different for different waves, but the velocit of the waves must all be the same 87. m/s, since it is determined b the tension and the linear mass densit of the string.) (.) Threshold of pain. Intensit of sound is 0. W/m at 7. m from a point source. At what distance from the point source will intensit reach the threshold of pain,.0 W/m? Denote the new distance D. Then.0 W/m πd = 0. W/m π (7. m) = power output of the point source. Hence D = 7. m (0. W/m /.0 W/m ) / =.87 m. (Note: πd is the area of a sphere of radius D.) Thus ou have to move 7. m .87 m =.0 m toward the source of sound to reach the threshold of sound. (.8) Interference of triangular pulses. v =.00 cm/s..00 cm.00 cm.00 cm.00 cm.00 cm.00 cm.00 cm.00 cm/s 0. s = 0. cm. The two pulses will barel touch after traveling for 0. s, producing:.00 cm.00 cm.00 cm.00 cm.00 cm.00 cm.00 cm/s 0.0 s =.0 cm. The two pulses now overlapped b.00 cm..00 cm.00 cm The superposed pulses now show a flat top for a width of.00 cm.
3 .00 cm/s 0.7 s =. cm. The two pulses now totall overlap. Their superposition gives one triangular pulse of twice the height, but the same old width of.00 cm. But this shape is maintained for onl one instant of time, since part of it is moving to the right, and part of it is moving to the left..00 cm.00 cm.0 cm.00 cm.00 cm.0 cm.00 cm/s.00 s =.00 cm. The two pulses have now moved pass each other b.0 cm. The superposition again produces the trapezoidal shape with a flat top of.00 cm just like when it has onl moved for 0. s. But notice the directions of the two horizontal arrows, which indicate the directions of motion of the two constituent pulses..00 cm.00 cm.00 cm/s. s =.0 cm. The two pulses have now moved pass each other b their own width, so the no longer overlap. The situation appears just like when the have moved for onl 0. s, except for the directions of the two horizontal arrows, which indicate the directions of motion of the two constituent pulses..00 cm.00 cm.00 cm.00 cm.00 cm.00 cm
4 (.) Distance between adjacent antinodes of a standing wave is.0 cm. A particle at an antinode oscillates in SHM with amplitude 0.80 cm and period s. The string is along x and is fixed at x = 0. (a) Adjacent nodes are also.0 cm apart. (b) Recall that cos (A B) + cos (A + B) = cos A cos B, which implies that A cos (kx ωt) + A cos (kx + ωt) = A cos (kx) cos (ωt). Hence the amplitude of a standing wave is a factor two larger than the amplitude of either of the two constituent traveling waves, but their wavelengths are the same, and their frequencies are the same. Thus: Wavelength of each traveling wave λ = 0.0 cm; Amplitude of each traveling wave A = 0. cm. Speed of each traveling wave v = λ f = λ / Τ = 0.0 cm / 0.07 s = 00 cm/s = m/s. (c) Maximum transverse speed of a point at the antinode of the standing wave = Aω = A (π / T) = 0.80 cm (π / s) = 7. cm/s. Minimum speed = 0. (d) Shortest distance bweteena node and an antinode is 7.0 cm. (.) Weightless ant. An ant of mass m stands on top of a horizontal stretched rope. Rope has mass per unit length μ and tension F. A sinusoidal transverse wave of wave length λ propagates on the rope. Motion is in the vertical plane. Find the minimum wave amplitude which will make the ant momentaril weightless. The mass m is so small that it will not affect the propagating wave. The ant will becomes momentaril weightless when it momentaril needs no support from the rope. This happens when its mass times the maximum downward acceleration of the SHM of the ant is exactl equal to its weight. This happens when the ant is at the highest point of its SHM. [Then just before or after this moment, the acceleration will still be downward but with a magnitude less than this maximum value, and the weight of the ant will need to be partiall cancelled b an upward support force from the rope, so that the net downward force is still equal to (now smaller) ma. Thus the weightlessness feeling happens onl at that brief moment when the ant is at the highest point of the SHM. During the half ccle of the motion when the acceleration is pointing upward, the weight is pointing in the wrong direction to provide the needed force. So an upward support force from the rope larger than the weight of the ant will exist, so that the net force is upward, and is still equal to ma. Thus in this half ccle the ant actuall weights more than its actual weight, if ou put a scale between the ant and the rope to measure this apparent weight.] The magnitude of the maximum acceleration of the SHM is mω. But ω = v k = (F / μ) (π / λ). Thus we must require ma(g / μ) (π / λ) = mg, giving A = (μ g / F) (λ / π). Let us check the unit. The unit of μ g is N/m. The unit of (μ g / F) is therefore just /m. The unit of (λ / π) is just m. So their product has the unit of m, which is the right unit for A.
5 (.6) More general sinusoidal wave: (x, t) = A cos (k x ω t + φ ). (a) At t = 0. (x, t = 0) = A cos (k x + φ ). For φ = 0, (x, 0) = A cos (kx): 0 x For φ = π/, (x, 0) = A cos (kx + π/): That is, even at x = 0, is alread cos (π/). 0 x For φ = π/, (x, 0) = A cos (kx + π/): That is, even at x = 0, is alread cos (π/). From this figure, we can see that it is the same as: (x, 0) = A sin (kx). The wave is then given b: (x, 0) = A sin (kx ωt). For φ = π/ = π/ + π/, (x, 0) = A cos (kx +π/): That is, even at x = 0, is alread cos (π/). 0 0 x x For φ = π/ = π + π/, (x, 0) = A cos (kx + π/): That is, even at x = 0, is alread cos (π/). From this figure, we can see that it is the same as: (x, 0) = + A sin (kx). The wave is then given b: (x, 0) = + A sin (kx ωt). 0 x
6 (b) Transverse velocit: v = / t = + Aω sin (k x ω t + φ ). (c) At t = 0, a particle at x = 0 has displacement = A /. Can one determine φ? No. There are several candidates: At t = x = 0, (0, 0) = + A cos (φ ). So we need to demand cos (φ ) = /. The answer is φ = π/ or 7π/. (Note: φ = 7π/ is the same as φ = π/, since their difference is π.) If we also know that a particle at x = 0 is moving toward = 0 at t = 0, determine φ. For x = 0, (0, t) = A cos ( ω t + φ ). We test φ = π/ and 7π/, and see which one works. We first calculate the transverse velocit v = / t = + Aω sin ( ω t + φ ), and evaluate it at t = 0, and get v (x = 0, t = 0) = + Aω sin (φ ). At φ = π/, this transverse velocit is positive, so it is not moving toward =0. So we reject it. At φ = 7π/, this transverse velocit is negative, so it is moving toward =0. So we accept this answer. Hence we conclude that φ = 7π/ (or, equivalentl, π/). Can one get this answer without doing so much math? es! First, one should realize that if (x, t) = A cos (kx ωt + φ), then the wave is moving toward positive x, or to the right in our plots. Then looking at our plot for φ = π/, and let the curve move to the right. One will find that will increase. at x = 0. So we should reject this answer. Then we do the same thing for φ = 7π/ or π/, and see that it can be accepted. (d) In general, to determine φ, we need to know: (i) (x = 0, t = 0); and (ii) the sign of v (x = 0, t = 0). (.68) Vibrating string. 0.0 cm long. Tension F =.00 N. Five stroboscopic pictures shown. Strobo rate is 000 flashes per minute. Maximum displacement occurs at flashes and, with no other maxima in between. P.0 cm.0 cm (a) Period: T = min = s (the time for 8 flashes). Frequenc f = / T = 0. Hz. Wavelength = 0.0 cm. All for either of the two the traveling wave on this string that are moving in opposite directions to form this standing wave. (b) The second harmonic (also known as the first overtone). (c) Speed of the traveling waves on this string v = λ / T = 0.0 cm / (0.096 s) = 0.8 cm/s =. m/s. (d) Point P. (i) In position it is moving with a transverse velocit of v = 0. (ii) In position it is moving with a transverse velocit of v = Aω = A(π f ) =
7 .0 cm (π 0. s ) =.0 cm 6.7 s = 96. cm/s =.96 m/s. (e) Mass of this string m = 0.0 m μ = 0.0 m [.00 N / (. m/s) ] = 0.08 kg = 8. g. (.7) String, Both ends held fixed. Vibrates in the third harmonic. Speed 9 m/s. Frequenc 0 Hz. Amplitude at an antinode is 0.00 cm. λ = 9 m/s / 0 Hz = 0.8 m. (a) Find amplitude of oscillation: (i) at x = 0.0 cm from the left end of the string. The general equation for the standing wave is : (x, t) = 0.00 m sin (π x / 0.8 m) sin (π 0 Hz t ) [We let x = 0 be the left end of the string. We should have the factor sin (π x / 0.8 m) because the left end is a node, and sin 0 = 0 Also, the wavelength is 0.8 m, and so at x = 0.8 m we should get sin (π) = 0.] So the amplitude of transverse oscillation at x = 0.0 cm is: 0.00 m sin (π 0 cm / 0.8 m) = 0. (ii) at x = 0.0 cm from the left end of the string, the amplitude is: 0.00 cm sin (π 0 cm / 0.8 m) = 0.00 cm. (iii) at x = 0.0 cm from the left end of the string, the amplitude is: 0.00 cm sin (π 0 cm / 0.8 m) = 0.8 cm/s. (b) At each point of part (a), find time to go from largest upward displacement to largest downward displacement. The answer is the same, and is equal to: T /, or / ( f ) = s, except at the nodes of the standing wave, i.e., at x = 0.0 cm in the three cases of part (a). (c) Maximum transverse velocit of the motion: v (x, t) = (x, t) / t = 0.00 cm (π 0 Hz) sin (π x / 0.8 m) cos (π 0 Hz t ) The amplitude of transverse velocit at distance x from the left end is 0.00 m (π 0 Hz) sin (π x / 0.8 m). So at x = 0.0 cm from the left end of the string, the maximum transverse velocit is: 0.00 m (π 0 Hz) sin (π 0.0 m / 0.8 m) = 0. at x = 0.0 cm from the left end of the string, the maximum transverse velocit is: 0.00 cm (π 0 Hz) sin (π 0.0 m / 0.8 m) = 60. cm/s = 6.0 m/s. at x = 0.0 cm from the left end of the string, the maximum transverse velocit is: 0.00 cm (π 0 Hz) sin (π 0.0 m / 0.8 m) = 6. cm/s =.7 m/s. Maximum transverse acceleration of the motion: a (x, t) = v (x, t) / t = 0.00 cm (π 0 Hz) sin (π x / 0.8 m) sin (π 0 Hz t ) The amplitude of transverse acceleration at distance x from the left end is 0.00 m (π 0 Hz) sin (π x / 0.8 m). So at x = 0.0 cm from the left end of the string, the maximum transverse acceleration is: 0.00 m (π 0 Hz) sin (π 0.0 m / 0.8 m) = 0. at x = 0.0 cm from the left end of the string, the maximum transverse acceleration is: 0.00 cm (π 0 Hz) sin (π 0.0 m / 0.8 m) =
8 60. cm/s = cm/s = 9096 m/s. at x = 0.0 cm from the left end of the string, the maximum transverse accelration is: 0.00 cm (π 0 Hz) sin (π 0.0 m / 0.8 m) = 67 cm/s = 6 m/s. Actuall, these answers are eas to see b looking at the figure: The second node from the left end is at 0.8 m, because ou see a whole wavelength between here and the left end. Then the first node from the left end must be at 0. m. The first antinode from the left end must be at 0. m. This is wh at x = 0. m we find the largest amplitudes of velocit and acceleration, and at x = 0. m we find vanishing amplitudes of velocit and acceleration. In addition, the amplitudes of velocit and acceleration at x = 0. m is down from those at x = 0. m b a factor sin (π/) = / = 0.707, as the xdependent factor is simpl a sine function to give sin (π/) = at x = 0. m, and to give sin (π) = 0 at x = 0. m. It will also give 0 at x = 0.8 m, and at x =. m, which is the right end of the string. The correspond to sin (π) and sin (π) from the xdependent factor.
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