Q1

Given the progressive wave equation{UTME 2008}

calculate the wavelength.

A. 12.4m B. 15.7m C. 17.5m D. 18.6m

Solution

using the wave equation

compare this wave equation with the one given in the question

K is the phase difference

B is the correct option

Q2

If a light wave has a wavelength of 500nm in air, what is the frequency of the wave?{UTME 2009}

A. 3.0 x 1014 Hz B. 6.0 x 1014 Hz C. 6.0 x 1012 Hz D. 2.5 x 1014 Hz

[c = 3 x 10^8 ms-1]

v = f**ℷ**

nanometer = 10^-9 m

3 x 10^8 = f x 500 x 10^-9

f = 3 x 10^8 / 500 x 10^-9

f = 6.0 x 10^14 Hz

B is the correct option

Q3

The wavelength of a wave travelling with a velocity of 420ms-1 is 42 m. What is its

period?{UTME 2010}

A. 1.0s B. 0.1s C. 0.5s D. 1.2s

Solution

v = **ℷ** / T

T = **ℷ** / v

T = 42 / 420

T = 0.1 s

B is the correct option

Q4

A microphone connected to the Y-plates of a cathode-ray oscilloscope (c.r.o.) is placed in front of a loudspeaker. The trace on the screen of the c.r.o. is shown below{Cambridge may/june 2016 p12}

The time-base setting is 0.5 ms cm–1 and the Y-plate sensitivity is 0.2 mV cm–1. What is the frequency of the sound from the loudspeaker and what is the amplitude of the trace on the c.r.o.?{ Cambridge may/june 2016 p12}

Solution

The y-sensitivity will measure the amplitide which is on the y-axis

The time-base setting will measure the period which is on the x-axis

From the graph representation a full cycle occupies 6boxes

and one box to the x-axis o.5ms

period (T) = no of boxes x value of one box(which is 0.5 ms)

T = 6 X o.5 = 3 ms

Frequency(f) = 1 /T

f = 1 / (3 x 10^-3) = 333 Hz = 330 Hz approximately

from the graph the amplitude occupies 3 boxes i.e. maximum displacement from the rest position

and one box to the y-axis o.2 mv

A mplitude = no of boxes x no of one box(which is 0.2mv)

A = 3 x 0.2mv

A = 0.6 mv

A is the correct option

Q5

The variation with time t of the displacement y of a wave X, as it passes a point P, is shown in

use the diagram to determine the frequency of wave X.{ Cambridge may/june 2016 p12}

from the graph the period is 4.0 ms

f = 1 / T

f = 1 / (4 x 10^-3)

f = 250 Hz