General Physics III - Practice Exam II Solutions - Fall ... - Amazon S3

General Physics III - Practice Exam II Solutions - Fall 20071. A string fixed at both ends is vibrating in a standing wave at its fundamental resonance frequency. Thestring vibrates according to the equationy(x,t) = 0.02 sin(πx/2)cos(20πt)where x and y are in meters and t is in seconds.(a) What is the wave speed for the string?(b) What is the length of the string?(c) If the string has a mass of 50g, what is the tension in the string?(d)(e)What is the maximum speed of a segment of the string?The vibrating string described above is placed over the open end of a tube filled with air, The opposite endof the tube is closed. The vibrating string generates a standing wave in the air column inside the tube at itsfundamental frequency. What is the length of the tube? (The speed of sound in air is 340 m/s.)

2AA bat flies towards a wall at a speed of 5.0 m/s. As it flies, the bat emits an ultrasonic sound wave withfrequency 30,000 Hz. What frequency does the bat hear in the reflected wave?Solution:f=fv + vv + vdetectorbat0 = f0=v − vsourcev − vbat30,890HzThe bat is a moving source and detector with the same speed. This is because the wall is stationary so thefrequency of the reflected sound is the same as the frequency that the fixed wall receives from the movingbat with the wave reflected in the opposite direction.Or one can solve this in two steps. First finding the freq f 1 that an observer at the wall would receivef 1 = f 0 (v / (v – v bat ))Then finding the frequency that the bat would hear from the sound wave reflectedfrom the wallf 2 = f 1 ((v + v bat ) / v) giving the same answer as above.2B. You have three tuning forks, A, B and C. Fork B has a frequency of 440 Hz. When A and B are soundedtogether, a beat frequency of 3 Hz is heard. When B and C are sounded together, the beat frequency is 4Hz.(a) What are the possible frequencies of A and C?(b) What beat frequencies are possible when A and C are sounded together?Solution:When A and B are sounded, we have⏐fA – fB⏐ = 3 Hz, sofA = fB ± 3 Hz = 440 Hz ± 3 Hz, so fA = 437 Hz, or 443 Hz.When C and B are sounded, we have⏐fC – fB⏐ = 4 Hz, sofC = fB ± 4 Hz = 440 Hz ± 4 Hz, so fC = 436 Hz, or 444 Hz.The possible beat frequencies when A and C are sounded are⏐fA – fC⏐ = ⏐437 Hz – 436 Hz⏐ = ⏐443 Hz – 444 Hz⏐ = 1 Hz, and⏐fA – fC⏐ = ⏐437 Hz – 444 Hz⏐ = ⏐443 Hz – 436 Hz⏐ = 7 Hz.

3.A. Two students are performing an experiment. Two springs are connected at a point J that isequidistant from the students' hands, as shown in the top-view diagram below. Student 2 creates a pulsehaving a displacement on only one side of the spring. A short time later, students 1 and 2 receive pulseson opposite sides of the spring.

4A. Two converging lenses (lens 1 and lens 2) with focal length f 1 = 20.0 cm and f 2 = 25.0 cm are placed20.0 cm apart along their common optical axis. An object is placed 60.0 cm in front of lens 1.Determine:(a) the position of the image from lens 1, if it is real or virtual, upright or inverted and its magnification;(b) the position of the final image from lens 2, if it is real or virtual, upright or inverted and its magnification.(c) Sketch a diagram showing the two lenses, the object (as an upright arrow), the image from lens 1and the final image.Solution:(a) The image from the first lens is:1 1 1 1 1 1q 1f 1p 120cm 60cm 30.0 cm⇒ q = + 30.0 cm.1The image is real since q 1 is + ; the image is inverted since m = - q 1 /p 1 is negativem = - (30.0 cm) / (60.0 cm) = -0.50(b) This image is the object for the second lens.Because it is beyond the second lens, it has a negative object distance:p 2= 20.0 cm – 30.0 cm = – 10.0 cm.We find the image formed by the refraction of the second lens:1 1 1+ =p2 q2 f2[1/(– 10.0 cm)] + (1/q 2) = 1/25.0 cm, which gives q 2= + 7.14 cm.The image is real since q 2 is +The total magnification is the product of the magnifications for the two lenses:m = m1m2 = (–q 1 /p 1 )(– q 2 /p 2 ) = q 1 q 2 / p 1 p 2= (+ 30.0 cm)(+ 7.14 cm)/(+ 60.0 cm)(– 10.0 cm) = – 0.357 ( image inverted since m is -).(c)OI 2 I F 1 2F 1F F21

4B. Violet light of wavelength 415 nm passes through a single narrow slit and creates a central diffraction peakwhich has a width (distance between the first minima on both sides of the central maximum) of 9.20 cm ona screen that is 2.55 m from the slit. How wide is the slit?Solution:tan θ1min = ½ (9.20 cm)/(255 cm) = 0.0180 = sin θ1min , because the angle is small.We find the slit width from:a sin θ1min = mλ;a (0.0180) = (1)(415 × 10 –9 m), which gives a = 2.30 × 10 –5 m =0.0230 mm.4C. How many lines per centimeter must diffraction grating have such that there will be no 2 nd order maximumfor light of any wavelength in the visible region (400 nm to 700 nm)?Solution:Because the angle increases with wavelength, to miss a complete order we use the smallest visiblewavelength. The maximum angle is 90°. We find the slit separation fromd sin θ = mλ;d sin 90° = (2)(400 × 10 –9 m), which gives d = 8.00 × 10 –7 m = 8.00 × 10 –5 cm.The number of lines/cm is1/d = 1/(8.00 × 10 –5 cm) = 12,500 lines/cm.