Waves Test

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Waves Test - 43

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Weekly Quiz Competition

Question 1
1 / -0

A string of mass $$m$$ and length $$l$$ is hanging from ceiling as shown in the figure. Wave in string move upward. $$v_{A}$$ and $$v_{B}$$ are the speed of wave at $$A$$ and $$B$$ respectively. Then $$v_{B}$$ is

A
$$\sqrt {3}v_{A}$$

B
$$v_{A}$$

C
$$< v_{A}$$

D
$$\sqrt {2}v_{A}$$

Solution

String mass $$=m$$
length $$=L$$
$$\mu =m/L$$
Tension at $$A={ T }_{ A }=\dfrac { m }{ 4 } \times g=mg/4$$
Velocity of wave at $$A={ V }_{ A }=\sqrt { \dfrac { { T }_{ A } }{ \mu } } =\sqrt { \dfrac { mg/4 }{ m/L } } =\sqrt { \dfrac { gL }{ 4 } } $$
Tension at $$B={ T }_{ B }=\dfrac { 3 }{ 4 } m\times g$$
speed of wave at $$B={ V }_{ B }=\sqrt { \dfrac { { T }_{ B } }{ \mu } } =\sqrt { \dfrac { 3mg }{ 4\mu } } =\sqrt { \dfrac { 3gL }{ 4 } } $$
$$\Rightarrow \quad \left[ { V }_{ B }=\sqrt { 3 } { V }_{ A } \right] $$

$$\therefore $$ Option (A) is correct.
Question 2
1 / -0

With propagation of longitudinal waves through a medium, the quantity transmitted is :

A
matter

B
energy

C
energy and matter

D
energy, matter and momentum

Solution

Hint :- Wave is the transmission of energy without transfer of matter.

Explanation:-
In propagation of longitudinal waves through a medium leads to transmission of energy through the medium without matter being transmitted. There is no movement of matter (mass).
Question 3
1 / -0

Two simple harmonic motions are represented by
$${ y }_{ 1 }=5\left( \sin { 2\pi t } +\sqrt { 3 } \cos { 2\pi t } \right) $$
$${ y }_{ 2 }=5\left( \sin { 2\pi t } +\cfrac { \pi }{ 4 } \right) $$
The ratio of the amplitude of two S.H.M's is

A
$$1:1$$

B
$$1:2$$

C
$$2:1$$

D
$$1:\sqrt { 3 } $$

Solution

Here, y1=5(sin2πt+3√cos2πt)
y2=5sin(2πt+π4)
as of the form of y1=αsin2πt+βcos2πt
Let α=rcosθ=5,β=rsinθ=53√
∴y1=rcosθsin2πt+rsinθcos2πt=rsin(2πt+θ)
also, α2+β2=r2cos2θ+r2sin2θ=r2
or r=α2+β2−−−−−−√=512+(3√)2−−−−−−−−−√=10
∴y1=10sin(2πt+θ)
∴A1A2=105=2
Question 4
1 / -0

A particle moves according to the law $$x = r\cos \dfrac {\pi t}{2}$$. The distance covered by it in the time interval between $$t = 0$$ and $$t = 3s$$ is

A
$$r$$

B
$$2r$$

C
$$3r$$

D
$$4r$$

Solution

Position of particle $$x = r\cos\dfrac{\pi t}{2}$$
At $$t = 0\ s$$, $$x = r\cos0 = r$$
So, at $$t=0 \ s$$, particle is at extreme position of SHM.
Time period of oscillation $$T = \dfrac{2\pi}{w}= \dfrac{2\pi}{\pi/2} = 4 \ s$$
In one time period, distance moved by particle $$ = 4r$$
So, in $$3 \ s$$ distance moved $$ = \dfrac{3}{4}\times 4r = 3r$$
Question 5
1 / -0

Two particle executing SHM of same frequency meet at $$x=+A/2$$, while moving in opposite direction. Phase difference between the particles is ($$A \rightarrow$$ Amplitude of both)

A
$$\cfrac{\pi}{6}$$

B
$$\cfrac{\pi}{3}$$

C
$$\cfrac{5\pi}{6}$$

D
$$\dfrac{-2\pi}{3}$$

Solution

For particle 1

$$x_1 = a.\sin (wt)$$ ...(1)

For particle 2

$$x_2 = a.\sin (wt+y)$$ ...(2)

For instance, $$x_1=x_2=\dfrac {a}{2}$$

From (1)

$$\dfrac {a}{2} = a.\sin (wt)$$

Solving we get,

$$wt=\dfrac{\pi}{6}$$ or $$wt=\dfrac{5\pi}{6}$$

Putting these value in (2),

$$\sin \dfrac{\pi}{6} = a.\sin \left ( \dfrac{5\pi}{6} +y\right ) = \dfrac {a}{2}$$

Solving this,

$$y=-\dfrac{2\pi}{3}$$
Question 6
1 / -0

Two wires made of the same material, one thick and the other thin, are connected to form a composite wire. The composite wire is subjected to some tension. A wave travelling along the wire crosses the junction point. The characteristic that doesn't undergoes a change at the junction point is

A
Frequency only

B
Speed of propagation only

C
Wavelength only

D
The speed of propagation as well as the wavelength

Solution

In case of composite wire:
1) frequency remains the same
2)wavelength change(decrease into thick wire because thick wire is a denser medium) because f=F
$$\cfrac{V_1}{\lambda_1} = \cfrac{V_2}{\lambda_2}$$
Also speed is decreased.
Question 7
1 / -0

Equations of a stationary wave and a travelling wave are $$y_1=1\,sin(kx)\,cos (\omega t)$$ and $$y_2=a\,sin\,(\omega t-kx)$$.The phase difference between two points $$x_1=\dfrac{\pi}{3k}$$ and $$x_2=\dfrac{3 \pi}{2k}$$ is $$\phi_1$$ for the first wave and $$\phi_2$$ for the second wave.The ratio $$\dfrac{\phi_1}{\phi_2}$$ is

A
1

B
5/6

C
3/4

D
6/7

Solution

Phase difference between two points in a standing wave =$$n\pi$$
Where n is number of nodes between two points.
Given points are $$x_1 = \cfrac{\pi}{3k} = \cfrac{60}{k}$$
$$x_2 = \cfrac{3\pi}{2k} = \cfrac{210}{k}$$
Equation of the standing wave
$$ y_1 = a \sin kx \cos \omega t$$
At node points $$ kx =n\pi$$
$$ x = \cfrac{n\pi}{k} \quad (n=0,1,2,3...)$$
So nodes are =$$ \cfrac{\pi}{k} , \cfrac{2\pi}{k} ....$$
$$ = \cfrac{180}{k} , \cfrac{360}{k} ....$$
Since there is only one node between phase difference $$ \phi_1 = \pi$$
For travelling wave $$ \phi _2 = \cfrac{2\pi}{\lambda} \triangle x$$
From the equation
$$y_2 = a \sin (\omega t - kx)$$
$$ k = \cfrac{2\pi}{\lambda}$$
$$ \therefore \phi_2 = k[x_2 - x_1] = k[\cfrac{3\pi}{2k} - \cfrac{\pi}{3k}] = \cfrac{7}{6}\pi$$
$$ \therefore \cfrac{\phi_1}{\phi_2} = \cfrac{\pi}{\cfrac{7}{6}\pi} = \cfrac{6}{7}$$
Question 8
1 / -0

How long after the beginning of motion is the displacement of a harmonically oscillating particle equal to one half its amplitude if the period is $$24\ s$$ and particle starts from rest

A
$$12\ s$$

B
$$2\ s$$

C
$$4\ s$$

D
$$6\ s$$

Solution

Let's draw the phasor diagram
$$\cos\theta=\cfrac{\cfrac{A}{2}}{A}=\cfrac{1}{2}\\ \theta 60°$$
On rotation of $$960°$$ it takes $$24$$ sec the for $$60°$$ rotation it will take $$\cfrac{24}{6}=4sec$$
Question 9
1 / -0

A string of length 'l' is fixed at both ends. It is vibrating in its 3^rd overtone with maximum amplitude 'a'.The amplitude at a distance l/3 from one end is?

A
$$14a$$

B
$$15$$ a

C
$$10$$ a

D
None of these

Solution

$$y=(a\sin \omega t)\cos kx\\y=(a\sin \omega t)\cos(\cfrac{2\pi}{l/2})x\\At \quad x=l/2\\y=(a\sin\omega t)\cos(\cfrac{2\pi}{l}\times2\times\cfrac{l}{3})\\y=(a\sin\omega t)\cos(\cfrac{4\pi}{3})$$
In this question, without calculating we can answer that none of these are correct.
Because given that maximum amplitude $$=a$$ then how can at any place amplitude will be greater than $$'a'$$. It is not possible.
Question 10
1 / -0

A particle executes SHM with time period $$T$$ and amplitude $$A$$. The maximum possible average velocity in time $${T}/{4}$$ is:

A
$$\cfrac{2A}{T}$$

B
$$\cfrac{4A}{T}$$

C
$$\cfrac{8A}{T}$$

D
$$\cfrac{4\sqrt {2}A}{T}$$

Solution

The maximum possible average velocity $$= \dfrac{\text{maximum displacement}}{\text{time}}$$
i.e. Vavg $$= \dfrac{dmax}{T/4}$$
here $$T \to 2\pi$$
$$\dfrac{T}{4} \to \dfrac{\pi}{2}$$
$$\therefore \theta = \dfrac{\pi}{4}$$
$$\therefore dmax = PQ = 2A\sin \theta$$
$$= 2a\sin \dfrac{\pi}{4}$$
$$\therefore dmax = \dfrac{2A}{\sqrt{2}} = \sqrt{2}A$$
$$\therefore Vavg = \dfrac{\sqrt{2}A}{T/4} = \dfrac{4\sqrt{2}A}{T}$$