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Question 566: [Waves]

(a) Source of sound has frequency f. Sound of wavelength λ is produced by the source.

(i) State

1. what is meant by frequency of the source,

2. distance moved, in terms of λ, by a wavefront during n oscillations of the source.

(ii) Use answers in (i) to deduce an expression for speed v of the wave in terms of f and λ.

(b) Waveform of a sound wave produced on screen of a cathode-ray oscilloscope (c.r.o.) is shown in Fig.

Time-base setting of the c.r.o. is 2.0 ms cm^–1.

(i) Determine frequency of the sound wave.

(ii) A second sound wave has same frequency as that calculated in (i). Amplitude of the two waves is the same but phase difference between them is 90°.

Reference: Past Exam Paper – June 2010 Paper 23 Q5


Solution 566:

(a)

(i)

1. The frequency of the source is the number of oscillations produced by the source per unit time

2.

{For 1 oscillation, the distance moved by a wavefront is λ}

Distance moved during n oscillations = nλ

(ii)

EITHER

Speed v = distance / time = (nλ) / t

But n / t = frequency f            hence Speed v= fλ

OR

If there are f oscillations per unit time, so (fλ) is the distance per unit time.

The distance per unit time is the speed v, so v = fλ

(b)

(i)

{1 period corresponds to 3 squares (3cm) on the screen. The time-base setting is such that 1cm represents 2ms.}

1.0 period is 3 × 2 = 6.0 ms

Frequency (= 1/ T) = 1 / (6 × 10^–3) = 170 Hz

(ii)

{1 complete wave (1 wavelength) corresponds to 360°. So, 90° would correspond to a quarter of a wavelength. So, the new wave should be displaced from the one shown by a distance equal to λ/4. Or, in terms of time, this is displaced by a time of T/4}

The wave should be drawn (with approx. same amplitude and) with correct phase difference

Question 567: [Matter > Density]
Diagram shows the atoms of a substance with atoms at the corners of a cube. Average separation of the atoms at a particular temperature is 15 nm.

When the temperature changes so that average separation becomes 17 nm, by which factor will the density of the substance change?
A 0.61                                     B 0.69                         C 0.78                         D 0.88

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q19



Solution 567:

Answer: B.

Density = mass / volume

The mass M is independent of changes in temperature,

Initial volume = 15 x 15 x 15 = 15³ nm³
Initial density ρ_i = M / (15³)

Final volume = 17 x 17 x 17 = 17³ nm³
Final density ρ_f = M / (17³)

Let the factor by which the density of the substance changes = y
ρ_f = y (ρ_i)
Factor y = ρ_f / ρ_i = [M / (17³)] / [M / (15³)] = 15³ / 17³ = 0.69











Question 568: [Nuclear Physics]
U⁺⁺ is doubly-ionised uranium atom. Uranium atom has a nucleon number of 235 and a proton number of 92.
In a simple model of the atom, how many particles are in this ionised atom?
A 235                          B 325                          C 327                          D 329

Reference: Past Exam Paper – June 2014 Paper 11 Q39



Solution 568:

Answer: B.

The nucleus of the uranium atom contains 235 nucleons (this includes the number of neutrons and protons).

A neutral atom would have 92 electrons (equal to the number of proton so that the atom is neutral), but this atom is doubly-ionised (2 electrons have been removed) and positive, so it must have 90 electrons.

There are (235 nucleons + 90 electrons =) 325 particles in total.

Question 569: [Kinematics]
Small steel ball falls freely under gravity after being released from rest.
Which graph best represents variation of the height h of the ball with time t?

Reference: Past Exam Paper – June 2010 Paper 12 Q8



Solution 569:

Answer: B.

Since the ball is released from rest, its height initially has greatest value. [D is incorrect]

The gradient of the graph represents the velocity of the ball.

At rest, the velocity is zero, so the gradient should be zero at t = 0. [C is incorrect]

Due to gravity, the ball accelerates as it falls. So, its speed increases with time. This is represented by a change in gradient in the graph. [A is incorrect]