1 The ionosphere contains free electrons. What is the amplitude of oscillation of the oscillation of these electrons when subject to a 200 kHz electromagnetic wave in which the oscillations of electric field have amplitude 5 ´ 10－3 V m －1 ?
A 3.2 ´ 10－15 m B 4.0 ´ 10－9 m C 2.5 ´ 10－8 m D 5.6 ´ 10－4 m
Force on electron F = eE = (1.6 ´ 10－19) (5 ´ 10－3)
Since Fnet = F = ma (m = 9.1 ´ 10－31 kg)
F = mw2x (w = 2pf)
Hence x = eE / (m4p2f2) = 5.6 ´ 10－4 m
Ans: D
2. A suspension bridge is to be built across a valley where it is known that the wind can gust at 5 s intervals. It is estimated that the speed of transverse waves along the span of the bridge would be 400 ms－1. The danger of resonant motions in the bridge at its fundamental frequency would be greatest if the span had a length of
A 2000 m B 1000 m C 400 m D 80 m
If l is the length of the suspension bridge and l is the wavelength associated with the speed of at which the wind is blowing, then
I = l/2 = 0.5 (v/f) = 0.5 (400) )5) = 1000 m
Ans: B
3.) A mass hanging from a light helical spring produces an equilibrium extension of 0.1 m. The mass is pulled vertically downwards by a distance of 0.02 m and then released. The angular frequency of the simple harmonic motion is 10 rad s m－1.The equation relating the displacement x of the mass from its equilibrium position and the time t after release is
A x/m = 0.02 sin [10(t/s)]
B x/m = 0.03 sin [10(t/s)]
C x/m = 0.02 cos [10(t/s)]
D x/m = 0.03 cos [10(t/s)]
Amplitude of simple harmonic motion, xo = 0.02 m since this is the distance from the equilibrium position. The motion results in a cosine graph since it starts from the maximum displaced position at t = 0.
Ans: C
4.) Values of the acceleration a of a particle moving in simple harmonic motion as a function of its displacement x are given in the table below.
| a / mm s－2 | 16 | 8 | 0 | －8 | －16 |
| x / mm | －4 | －2 | 0 | 2 | 4 |
The period of the motion is
A 1/p s B 2/p s C p / 2 s D p s
Angular frequency, w2 = －a/x = －16/(－4) = 4 or w = 2 rad /s
Hence, period of the motion , T = 2p/w = p s
Ans: C
5 In which group below do all three quantities remain constant when a particle moves in simple harmonic motion?
| A | acceleration | force | total energy |
| B | force | total energy | amplitude |
| C | total energy | amplitude | angular frequency |
| D | amplitude | angular frequency | acceleration |
In simple harmonic motion, the energy is conserved and hence the total energy (sum of KE and PE) remains constant. The equation that describes the motion in terms of displacement x is given by
a = －w2x
Amplitude of the SHM is the maximum displacement of the motion and hence remains unchanged.
Ans: C
6.) A particle P performs simple harmonic motion of period 8 s about a fixed point O in the horizontal direction. The maximum displacement of P from O is 5 m. If P is initially at O and moving to the left, the 17 s later P will be
A moving towards O.
B moving with increasing speed.
C moving with increasing acceleration.
D at a distance of 2.5 m to the left of O.
P passes by O at 0, 8 and 16 s. At t = 17 s, it will be moving leftward away from O. Since a µ x, acceleration increases as displacement from O increases.
Ans: C
7 A point mass moves with simple harmonic motion. Which of the following statements is false?
| A | The maximum kinetic energy is dependent on the frequency of the oscillation. |
| B | The time taken for the system to change from maximum kinetic energy to maximum potential energy is a quarter of the period of the oscillation. |
| C | An oscillation system with larger amplitude will have a greater maximum velocity |
| D | An oscillation system with larger amplitude will have longer period. |
Ans: D
During the “live” telecast of Campus mega hotstar, a television viewer in Pasir Ris hears the sound picked up by a microphone directly in front of a contestant. This viewer is seated 2.00 m from the television set.
A “live” audience is in the recording studio located 10.0 m from the microphone. The “live” audience and the television viewer hear the same sound at a time difference of 0.025 s.
If the distance between the “live” audience and the television viewer is 15.0 km and assuming that the electrical signal is transmitted via land cables at the speed of light, determine the speed of sound.
A 311 ms-1 B 319 ms-1 C 330 ms-1 D 343 ms-1
t1 = time taken for sound travel from mic to live viewer
t2 = time taken for sound travel from mic to television
t3 = time taken for sound travel from television to home viewer
Assuming that home viewer hears the sound first,
t1 -(t2 + t3) = 0.025
10/vs -(15´103/c + 2/vs) = 0.025 where (c=3.0´108 ms-1)
vs = 319 ms-1
Ans: B
9)
A small loudspeaker at P generates a sound wave of frequency 660 Hz. If the speed of sound in air is 330 ms-1, the phase difference between the air vibrations at Q and R, 0.25 m apart, is
A zero B 2p radians C p/2 radians D p radians
v=fl
l=330/660=0.50 m
Df=o.25/0.50 ´ 2p = p rad
Ans: D
The intensity of a wave is proportional to the square of the amplitude of the wave. If two waves of the same frequency and different amplitudes are superimposed at a point in phase the total intensity at that point is proportional to
A the sum of the intensities of the two waves
B the mean value of the intensities of the two waves
C the square of the sum of the two amplitudes
D the square of the mean value of the two amplitudes
Ans: C
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