# Physics 9702 Doubts | Help Page 55

__Question 318: [Waves > Phase difference]__Can you please explain how to draw two waves with a phase difference? For example, "

*draw two waves of identical frequency, wavelength and amplitude and with a phase difference of 5π/8*".

__Solution 318:__Consider the wave drawn (in red). It has a wavelength of 12cm.

We want to draw a wave that has a phase difference of 5π/8 to the wave in red. This new wave will be in green.

Note that one wavelength corresponds to a phase difference of 360

^{o}(that is the wave would be exactly in the same position).

First, we find the equivalence of 5π/8 in degrees. (or you may work in radians itself if you are more familiar with it)

5π/8 = 5 (180) / 8 = 112.5

^{o}

So, the wave to be drawn has a phase difference of 112.5

^{o}. Now, we need to calculate how much of the length of a wave this would correspond.

360

^{o}corresponds to 12cm

112.5

^{o}corresponds to 12 x (112.5 / 360) = 3.75cm

Point A labeled on the wave above is at a distance of 3.75cm from the start of the red wave (from point O).

Therefore, if the red wave starts (point O) at position 1 indicated below, the green wave would have point A at position 1.

The green wave starts with point A at position 1 and the wave continues with a similar shape as the red wave beyond point A.

In the above diagram, the 2 waves have a phase difference of 5π/8 rad.

Note that the red wave could also have started differently (with a point other than point O at position 1). Let’s say the red wave starts at point X. Then, the green wave would have to start with a point (on the wave) being at a distance of 3.75m from point X. The 2 waves would still have a phase difference of 5π/8 rad.

If we want 2 waves with a different phase difference, we need to re-calculate the corresponding difference in positions of the points (as done above).

__Question 319: [Physics of Materials ~ Matter > Young modulus]__A uniform, vertical wire is stretched by hanging a mass from its lower end. Which of the following does not affect the strain in the wire?

A the stress

B its unstretched length

C its cross-sectional area

D the load applied

E the Young modulus of the metal

**Reference:**

*Past Exam Paper – J90 / I / 23*

__Solution 319:__**Answer: B.**

As a result of the wire being stretched, it is under strain.

The strain produced in a wire is proportional to the stress within the limit of proportionality. Within this limit, the ratio of stress / strain is a constant whose value depends on the material of the wire; and is known as the

**Young modulus**of the material.

Therefore, the strain depends on the stress applied to the wire. [A is incorrect]

Stress = Force / Cross-sectional Area. [C and D are incorrect]

The Young modulus is a property of the material of the wire and within the limit of proportionality, the ratio stress / strain = Young modulus. So, the strain also depends on the Young modulus of the metal. [E is incorrect]

The original length does not affect the strain in the wire. Strain = extension / original length. But the extension depends on the force applied. The original length does not affect the extension.

__Question 320: [Forces > Hooke’s law]__Some weight-lifters use a ‘chest expander’, consisting of a strong spring with a handle at each end, to exercise chest and arm muscles.

For one such chest expander, the relation between the force F applied by the weight-lifter and the extension x of the spring is shown in the graph.

Which graph best shows how W, the work done, depends on the extension x?

**Reference:**

*Past Exam Paper – N93 / I / 21*

__Solution 320:__**Answer: D.**

From the force-extension graph, it is seen that the spring obeys Hooke’s law [F = kx].

Work done, W = ½ Fx = ½ kx

^{2}

Work done W depends on x

^{2}.

It cannot be linear. [A, B and C incorrect]

Graph D represents a dependence on x

^{2}. (Consider the shape of the graph of y = x

^{2})

__Question 321: [Dynamics > Momentum]__A stationary body explodes into two components of masses m and 2m.

The components gain kinetic energies X and Y respectively.

What is value of the ratio X/Y?

A ¼ B ½ C 2/1 D 4/1

**Reference:**

*Past Exam Paper – June 2009 Paper 1 Q10 & June 2014 Paper 13 Q12*

__Solution 321:__**Answer: C.**

From the conservation of momentum, the sum of momentum before the explosion should be equal to the sum of momentum after the explosion.

Before explosion, the body was stationary. So, its momentum is zero.

Momentum of component 2m + Momentum of component m = 0

Momentum of component 2m = – Momentum of component m

Therefore, the magnitudes of momentum of the components are equal.

Let the velocity of component of mass 2m be v.

Magnitude of momentum of component 2m = Magnitude of momentum of component m

(2m) v = m (velocity)

Velocity of component of mass m = 2mv / m = 2v

Kinetic energy = ½ (mass) (velocity)

^{2}

X = 0.5 (m) (2v)

^{2}= 2mv

^{2}

Y = 0.5 (2m) (v)

^{2}= mv

^{2}

Ratio X / Y = 2

__Question 322: [Matter > Strain energy]__Graph shows the non-linear force-extension curve for wire made from a new composite material.

What could be the value of strain energy stored in the wire when it is stretched to point P?

A 0.09 J B 0.10 J C 0.11 J D 0.20 J

**Reference:**

*Past Exam Paper – November 2009 Paper 12 Q20*

__Solution 322:__**Answer: C.**

The strain energy is given by the
area under the force-extension curve.

Consider a

__straight__line joining point (0, 0) and point P (2.0, 100).**This line would be below the curve shown in the question.**
For the linear force-extension curve
(drawn joining point (0, 0) to point P),

Strain energy = ½ Fx = ½ (100)(2x10

^{-3}) = 0.1J.
But the shape of graph implies that
the area of under the curve shown in the question should be greater than the
area calculated. [A and B are incorrect]

However, the answer should be about 10%
greater than this straight line graph value to account for the difference. Choice
D (0.20 J) is twice the value calculated. This is too big. [D is incorrect]

__Question 323: [Physics of Materials ~ Matter > Young modulus]__The stress σ required to fracture a solid can be expressed as

σ = k (γE / d)

where k
is a dimensionless constant, E is the Young modulus, and d is the distance
between planes of atoms separated in fracture. Which one of the following
quantities could γ be to make the equation dimensionally consistent? ^{½}A an energy per unit area

B a force per unit area

C a force

D an energy

E a density

**Reference:**

*Past Exam Paper – J79 / II / 40*

__Solution 323:__**Answer: A.**

Stress σ = Force / Area

Units of stress: Nm

^{-2}

Consider the units of the right-hand side.

Young modulus E = stress / strain

Strain (= extension / original length) has no units

Units of E = Nm

^{-2}

Units of (γE / d)

^{½}= {[γ] Nm

^{-2}m

^{-1}}

^{0.5}= {[γ] Nm

^{-3}}

^{0.5}

For the equation to be dimensionally consistent, the overall units on the right-hand side should be equal to those on the left-hand side.

{[γ] Nm

^{-3}}

^{0.5}= Nm

^{-2}

{Square both sides,}

[γ] Nm

^{-3}= N

^{2}m

^{-4}

[γ] = N

^{2}m

^{-4}N

^{-1}m

^{3}= Nm

^{-1}

Consider choice A:

Unit of energy per unit area = Jm

^{-2}= (Nm) m

^{-2}= Nm

^{-1}

So, choice A is the correct answer. [The others may be checked and found to be incorrect]

2/M/J/02 Q.8(a)

ReplyDeleteThis question is explained as Solution 327 at

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how to find phase difference when two waves are given on the same graph Maj june 2015 q24 variant 12

ReplyDeleteSee solution 1063 at

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