MCAT Physics Problem Solving Drill 1: Waves and Periodic Motion Question No. 1 of 10 Question 1. Two waves on identical strings have frequencies in a ratio of 3 to. If their wave speeds are the same, then how do their wavelengths compare? Question #01 (A) 3: (B) :3 (C) 9: (D) :9 A. Incorrect! Frequency and wavelength are not directly proportional to each other. B. Correct! Frequency and wavelength are inversely proportional to each other. The wave with the greatest frequency has the shortest wavelength. No, square of the frequency can never be proportional to wavelength. D. Incorrect! This brings a crude and complex relation between frequency and wavelength. Frequency and wavelength are inversely proportional to each other. The wave with the greatest frequency has the shortest wavelength. Since the period is the reciprocal of the frequency the expression 1/f can be substituted into the above equation for period. Rearranging the equation yields a new equation of the form: Speed = Wavelength * Frequency The above equation is known as the wave equation. It states the mathematical relationship between the speed (v) of a wave and its wavelength (λ) and frequency (f). Using the symbols v, λ, and f, the equation can be rewritten as v = f * λ Frequency and wavelength are inversely proportional to each other
Question No. of 10 Question. An ocean wave has amplitude of.5 m. Weather conditions suddenly change such that the wave has amplitude of 5.0 m. The amount of energy transported by the wave is. Question #0 (A) Halved (B) Doubled (C) Quadrupled (D) Remains the same A. Incorrect! No, the energy E is proportional to square of the amplitude not the other way. B. Incorrect! Oh! Energy and amplitude never bear a direct proportionality. C. Correct! The energy transported by a wave is directly proportional to the square of the amplitude. Energy α (amplitude ) from the relation E = ½ k r D. Incorrect! How can the same energy be transported don t you think the energy should increase when amplitude of the wave increases commonsense isn t it? Answer :C (quadrupled) The energy transported by a wave is directly proportional to the square of the amplitude. Energy α (amplitude ) from the relation E = ½ k r So whatever change occurs in the amplitude, the square of that effect impacts the energy. When amplitude is doubled the energy increases by four time Equations are guides to thinking about how an impact on one variable affects another.
Question No. 3 of 10 Question 3. The time required for waves moving with a velocity v = 340 m/s to travel from the tuning fork to point P is. Question #03 (A) 8.5 sec. (B) 0.011 sec. (C) 0.19 sec. (D) 0.11 sec. A. Incorrect! t = d/v and not v/d B. Incorrect! Check the calculation. It s not 40/3400. Your calculation is not correct. D. Correct! GIVEN: v = 340 m/s, d = 40 m and f = 1000 Hz - To find Time frequency is not required. Answer: D GIVEN: v = 340 m/s, d = 40 m and f = 1000 Hz - To find Time frequency is not required. To find time Use v = d / t implies t = d / v t=d/v =(40/340) (sec) = 0.11sec
Question No. 4 of 10 Question 4. The factors affecting the speed of a wave are (speed of the wave is effected by). Question #04 (A) The properties of the medium through which the wave travels (B) The wavelength of the wave (C) The frequency of the wave (D) Both the wavelength and the frequency of the wave. A. Correct! The speed of the wave depends upon the properties of the medium. It is independent of wavelength and frequency. B. Incorrect! The speed of the wave does not depend on the wavelength of the wave. Wavelength is a characteristic of a wave and not the propagation of the disturbance. The speed of the wave does not depend on the frequency of the wave. Frequency is the occurrences of the disturbance in a unit time. D. Incorrect! If Answers B and C are incorrect then how can this be a relevant answer? Answer: A Whenever the medium is the same, the speed of the wave is the same. However, when the medium changes, the speed changes. The speed of the wave depends upon the properties of the medium.
Question No. 5 of 5 Question 5. What is the he number of nodes in the standing wave shown in the diagram? Question #05 (A) 15 (B) 13 (C) 8 (D) 14 A. Correct! There are 15 points where the resultant displacement of the standing waves pattern is zero. Hence, the number of nodes is 15. B. Incorrect! It s a question of who counts the zero disturbance points precisely. If you got it as 13, then to whom will you give the remaining two? What is the question read carefully count the number of nodes and not number of complete waves. D. Incorrect! It s again a question of who counts the zero disturbance points precisely. If you got it as 14, then you have a starting or ending count wrong. 1 3 4 5 6 7 8 9 10 11 1 13 14 15 Answer: A(15 nodes) There are 15 positions along the medium which have no displacement. Do not commit the common mistake of ignoring the end point nodes
Question No. 6 of 10 Question 6. A mass on a spring oscillates with a period of 4 s. If the mass is quadrupled, the new period is: Question #06 (A) s (B) 4 s (C) 8 s (D) 16 s A. Incorrect. The period will increase, not decrease. Consider the angular frequency of a simple harmonic oscillator. B. Incorrect. The period will not remain the same. Consider the angular frequency of a simple harmonic oscillator. C. Correct! The period is given by π (m/k). Increasing mass by a factor of 4 leads to an increase by a factor of 4=. See solution below. D. Incorrect. Although the period will increase, this is not the correct answer. The angular frequency, w, of a simple harmonic oscillator is given in terms of the spring constant, k, and the mass, m. k ω= m. The period, T, is related to the angular frequency as follows: π T= ω so the period can be written in terms of the spring constant m T=π k Denote the new period by T. The new period is: T'=π 4m = π k m k The new period can be written in terms of the old period as: T'=T Thus the period doubles, so (C) is the correct answer.
Question No. 7 of 10 Question 8 What is the frequency of a 30 cm long pendulum? The pendulum is a 100 gram mass whose center is attached to a 30 cm string that is anchored at one end. It is free to swing and vibrate like a pendulum. Question #07 (A) 1.1 Hertz (B) 110 Hertz (C) 0.091 Hertz (D) 0.91 Hertz A. Incorrect! This is the period of the swinging pendulum. However, in this case, the frequency is asked. Use the relationship between frequency and period. B. Incorrect! You may have used the mass of the pendulum to get the answer. The mass doesn t affect the period or frequency at all. You forgot to change the length unit into meters to match the acceleration of gravity units. D. Correct! First find the period of the pendulum. Be sure to convert the given length unit into meters. Then take the reciprocal to get the frequency. Note that the mass isn t used at all. Use the formula for the period of a simple pendulum. T = π l g Since we will input the acceleration due to gravity as 9.8m/s/s, we should also have our pendulum length in meters. 1m 30cm x =.30m 100cm Substitute this value, the acceleration due to gravity, and pi into the formula..30m T = (3.14) 1.1seconds 9.8m/s = This is the period of the pendulum. Period and frequency are reciprocals. f = 1/T = 1/1.1s =.91 Hertz Note that the given mass of 100 grams is irrelevant. The mass of a pendulum does not alter or affect its period or frequency at all. Question No. 8 of 10
Question 8. A simple harmonic oscillator s position is given by x(t)=a sin(t + 30 ). For what value of t is kinetic energy of the oscillator greatest? Question #08 (A) -15 (B) 15 (C) 30 (D) 45 A. Incorrect. At this value of t, the oscillator s position would be at x = 0, so the KE would be zero; obviously this value would not maximize the KE. B. Incorrect. This value would make the argument (the quantity inside the parenthesis of the sine function) have a value of 60 ; the sine function is maximized when its argument is 90. C. Correct! The argument of the sine function must be 90 to maximize the kinetic energy, which is equal to ½kx (t). D. Incorrect. Consider the effect of the phase of 30 ; the KE of the oscillator would not be greatest at ωt = 90 but ωt = 60. The kinetic energy, KE, is given by 1 1 ( ) KE= kx (t)= ka sin t+30. This function is maximized when the sine term is equal to 1, or when the argument of the sine term is 90. Solve the following equation to find t. sin( t+30 ) =1 t+30 =90 t=60 t=30. So (C) is the correct answer.
Question No. 9 of 10 Question 9. A 0.50 kg mass is attached to a spring and released. It is observed to make 10 complete vibrations in 13.0 seconds. What is the spring constant, or force constant, of that particular spring? Question #09 (A) 33.5 kg/s² (B) 11.7 kg/s² (C) 1.85 kg/s² (D) 0.11kg/s² A. Incorrect! You inverted the vibrations and time measurements when finding the period. Period is time/1 wave. B. Correct! Use the time per 10 waves to find the period, time per 1 wave. Then use this value in the formula for the period of an oscillating spring. Solve for the spring constant, k. You made a mathematical error when rearranging the period formula. When solving for k, be sure to square the and pi. D. Incorrect! The time given is not the period of the oscillation. It is the time for 10 complete oscillations. To get the period, divide this given time by 10. This gives the time per wave. This should be used with the formula to find the period of an oscillating mass. Use the formula for the period of an oscillating spring. T = π m k In this case we aren t solving for the period. That can be obtained from more direct measurements. It was observed that the oscillating mass took 13.0 seconds for 10 complete vibrations. Thus, 13.0s/10 vibrations = 1.3 sec/vibration. This is the period, T. We can rearrange the period formula to solve for k, spring constant. m( π) k = T Substituting the known values give:.50kg( π) k = = 11.7kg/s (1.3s) Notice the unusual units, this is also equal to N/m which may be more familiar. Question No. 10 of 10
Question 10. A mass attached to a spring undergoes simple harmonic motion. At what point is the velocity equal to zero? Question #10 (A) At the equilibrium position (B) At the points of maximum displacement (C) At an intermediate point between the equilibrium position and the maximum displacement point (D) None of the above A. Incorrect. At the equilibrium position, the kinetic energy attains a maximum value, so the velocity is maximal and non-zero. B. Correct! At the point of maximum displacement, the potential energy is at maximum; hence the velocity has a minimum value of zero. C. Incorrect. At an intermediate point, some of the energy will be in kinetic energy, and some of the energy will be in potential energy. The kinetic energy will not be maximized, and neither will the velocity. D. Incorrect. One of the above choices is correct, so discard this choice. The total energy is given by: 1 1 1 E tot=ke+pe= mv + kx = kx m. Where x m represents the maximum displacement. From the previous equation, the velocity squared is given by: k v= x -x ( m ) m So we see when x=x m, v=0. Thus at the points of maximum displacement, the particle s velocity approaches zero. So (B) is the correct answer. You can also visualize an oscillating, or bouncing, spring. Only at the very ends of its motion will it mometarily stop before going in the other direction.