(a) Explain; (i) wave motion (ii) stationary wave (b)(i) List four physical properties of a wave (ii) Define amplitude and use it to distinguish between the...

Assessment: WAEC SSCE - Physics - 2004 (Essay) Subject: Physics

Question 1 Report

(a) Explain;

(i) wave motion

(ii) stationary wave 

(b)(i) List four physical properties of a wave

(ii) Define amplitude and use it to distinguish between the node and antinode of a stationary wave

(iii) List the factors on which the frequency of vibration in a stretched string depends

(c) The equation, y = 5 sin (3x - 4t), where y is in millimeters, x is in meres and t is in seconds represents a wave motion . Determine the;

(i) frequency

(ii) period

(iii) speed of the wave

 

 

 

 

 

 

Answer Details

(a)(i) Wave motion: a disturbance that travels through a medium (or space) transferring energy from one point to another without any bulk transfer of the particles of the medium; the particles merely vibrate about their fixed mean positions.

(ii) Stationary (standing) wave: a wave formed when two progressive waves of the same frequency and amplitude travelling in opposite directions superpose; it has fixed points of zero amplitude (nodes) and points of maximum amplitude (antinodes), and the wave profile does not advance.

(b)(i) Four physical properties of a wave: amplitude, wavelength, frequency (or period), and speed (velocity).

(ii) Amplitude is the maximum displacement of a particle from its equilibrium (mean) position. In a stationary wave, a node is a point of zero amplitude (the particles do not move), while an antinode is a point of maximum amplitude.

(iii) The frequency of vibration of a stretched string depends on its length, the tension in the string, and its mass per unit length (linear density).

(c) Comparing \(y = 5\sin(3x - 4t)\) with \(y = A\sin(kx - \omega t)\): \(A = 5\ \text{mm}\), \(k = 3\ \text{m}^{-1}\), \(\omega = 4\ \text{rad s}^{-1}\).

(i) Frequency:

\[ f = \frac{\omega}{2\pi} = \frac{4}{2\pi} = 0.637\ \text{Hz} \]

(ii) Period:

\[ T = \frac{1}{f} = \frac{2\pi}{\omega} = \frac{2\pi}{4} = 1.57\ \text{s} \]

(iii) Speed:

\[ v = \frac{\omega}{k} = \frac{4}{3} = 1.33\ \text{ms}^{-1} \]

(Equivalently \(v = f\lambda\), with \(\lambda = \tfrac{2\pi}{k} = 2.09\ \text{m}\).)

Download The App On Google Playstore

Everything you need to excel in your exams

Green Bridge CBT Mobile App
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning