HCI 09 Waves Solutions (Self-Review Questions)

Topic 9: Waves Motion Suggested Solutions to Self-Review Questions S1 Answer: B Since the speed is independent of frequency  Speed is constant 1vv f f f      When 0, f  When , f0 [You should be familiar some of the common graphs and their features. This will help you do the questions pertaining to graphs esp. in MCQ Some common graphs: y = mx; y = mx+c, c 0 ; y = Ax2 ; y = 1/x ; y = x ; y = ln x ; y = ex ; y = e-x.] S2 Answer: B 112 10 cm 2.00 ms cm2 T  T = 8.00 ms f = 1/T = 1/(8.00 ms) = 125 Hz S3 Answer: C The diagram indicates that the wavelength (distance between consecutive rarefactions/ compressions) is 8 m. 12 1.5 Hz8 vf vf        S4 Answer: D (recall that any point on the string is undergoing SHM) Statement A: False. Speed of the element at P is zero, as it is at its highest point of motion. Statement B: False. This is a travelling wave, so all points on the wave are in SHM and hence displacement of the element at Q varies sinusoidally with time. Statement C: False. Each element of the wave is performing the same SHM with the same amplitude. So the (total) energy of each element is the same. Statement D: True. Since every element on the wave is in SHM, a = -2x. For S, the displacement is maximum, therefore the magnitude of a is also maximum. The negative sign of the acceleration indicates the direction of the acceleration which is pointing towards the equilibrium position i.e. in the negative direction. S5 c = 3.0  108 m s-1 Since f = c  , hence, maximum frequency corresponds to smallest wavelength. Max f = 8 14 9 3.0 10 7.5 10 Hz(400 10 )  

S6 Answer: C Students are expected to memorise the sequence of EM wave spectrum as well as the estimated corresponding wavelength range. S7 From the 1st graph, 3T = 14.5 ms  T = 4.83 ms From the 2nd graph, 3 2.7 m2   = 1.80 m Therefore, v = f = 3721083.4 80.1 3  T  m s-1 S8 Answer: B Using 1 (0.20 m)2 rad320 m s22 () 400 Hz x        . S9 Answer: C It is given that both waves have the same frequency, hence the same period. 2 /4 2 rad2 t T T T          . One is a sine wave and the other one is a cosine wave lagging behind by a quarter of period. Hence, the phase difference between the waves must be 90° or π/2 rad. Students chose B or D must have erroneously focused on the intersection points between the waves. T T/4

S10 (a) Intensity  2 1 r where r is the distance from the source 2 21 12   Ir Ir  I2 = 2 2 24.3 m( ) 0.026 W m 0.0500 W m3.1 m  (b) Energy = <P>  time = Intensity  Area  time = (0.026 W m-2)  (4 )  (4.3 m)2  (60  60)s = 2.17  104 J S11 Answer: A Intensity  A2f2 where A is the amplitude and f is the frequency. 22 22 2 2 2 1 1 1 0.5 cm 6.0 Hz 1.5 cm 3.0 Hz I A f I A f             