Waves Test

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Waves Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

A particle performing S.H.M. starts from equilibrium position and its time period is $$16$$ second. After $$2$$ seconds its velocity is $$\pi \ m/s$$. Amplitude of oscillation is $$(\cos 45^{\circ} = \dfrac {1}{\sqrt {2}})$$

A
$$2\sqrt {2} m$$

B
$$4\sqrt {2}m$$

C
$$6\sqrt {2} m$$

D
$$8\sqrt {2}m$$

Solution

let $$x=asin\omega t$$
on differentiating both side $$v=A\omega cos\omega t$$
$$\pi=A\times\dfrac{2\pi}{T}cos\dfrac{2\pi}{T}t$$
on putting T=16 sec and t=2 sec
$$8=Acos45$$
$$A=8\sqrt{2}$$
Question 2
1 / -0

Two sound waves having a phase difference of $$60^o$$, have a path difference of :

A
$$\dfrac{\lambda}{3}$$

B
$$\dfrac{\lambda}{6}$$

C
$$\dfrac{\lambda}{9}$$

D
$$\lambda$$

Solution

Path difference $$\Delta = \dfrac{\lambda}{2\pi}\times \Delta \phi$$
$$=\dfrac{\lambda}{2\pi}\times \dfrac{\pi}{3} = \dfrac{\lambda}{6}$$
Question 3
1 / -0

A progressive wave is represented by y = 12 sin (5t - 4x) cm. On this wave, how far away are the two points having phase difference of 90$$^o$$?

A
$$\dfrac{\pi}{2} cm$$

B
$$\dfrac{\pi}{4} cm$$

C
$$\dfrac{\pi}{8} cm$$

D
$$\dfrac{\pi}{16} cm$$

Solution

Given equation for the wave is $$y = 12 \textrm{sin}(5t-4x)$$

Phase is given by $$5t-4x$$. At a given instant, let two points $$x_1$$ and $$x_2$$ differ in phase by $$90^o = \pi/2$$.
$$\Rightarrow (5t-4x_1)-(5t-4x_2) = \pi/2$$
$$\Rightarrow 4|x_2-x_1| = \pi/2$$

Thus, distance between the two points is $$\pi/8$$ $$\textrm{cm}$$
Question 4
1 / -0

A simple harmonic wave of amplitude $$8$$ unit travels along positive $$x$$-axis. At any given instant of time, for a particle at a distance of $$10 cm$$ from the origin, the displacement is $$+6$$ unit and for a particle at a distance of $$25 cm$$ from the origin, the displacement is $$+4$$ unit. Calculate the wavelength.

A
$$200 cm$$

B
$$230 cm$$

C
$$210 cm$$

D
$$250 cm$$

Solution

$$Y=A\sin { \dfrac { 2\pi }{ \lambda } } \left( vt-x \right)$$
$$ \dfrac { Y }{ A } =\sin { 2\pi } \left( \dfrac { t }{ T } -\dfrac { x }{ \lambda } \right)$$
In first case, $$ \dfrac { { Y }_{ 1 } }{ A } =\sin { 2\pi } \left( \dfrac { t }{ T } -\dfrac { { x }_{ 1 } }{ \lambda } \right) $$
Here, $${ Y }_{ 1 }=+6$$, $$A=8$$, $${ x }_{ 1 }=10 cm$$
$$\dfrac { 6 }{ 8 } =\sin { 2\pi } \left( \dfrac { t }{ T } -\dfrac { 10 }{ \lambda } \right) $$ .....(i)
Similarly in the second case,
$$\dfrac { 4 }{ 8 } =\sin { 2\pi } \left( \dfrac { t }{ T } -\dfrac { 25 }{ \lambda } \right)$$ .....(ii)
From equation (i),
$$ 2\lambda \left( \dfrac { t }{ T } -\dfrac { 10 }{ \lambda } \right) =\sin ^{ -1 }{ \left( \dfrac { 6 }{ 8 } \right) } =0.85 rad$$
$$\Rightarrow \dfrac { t }{ T } -\dfrac { 10 }{ \lambda } =0.14$$ ....(iii)
Similarly from equation (ii),
$$2\pi \left( \dfrac { t }{ T } -\dfrac { 25 }{ \lambda } \right) =\sin ^{ -1 }{ \left( \dfrac { 4 }{ 8 } \right) } =\dfrac { \pi }{ 6 } rad$$
$$\Rightarrow \dfrac { t }{ T } -\dfrac { 25 }{ \lambda } =0.08$$ ....(iv)
Subtracting equation (iv) from equation (iii), we get
$$\dfrac { 15 }{ \lambda } =0.06$$
$$\Rightarrow \lambda =250 cm$$
Question 5
1 / -0

The equation of a simple harmonic wave is given by $$y = 6 \sin{2\pi} \left(2t-0.1x\right)$$, where $$x$$ and $$y$$ are in $$mm$$ and $$t$$ is in seconds. The phase difference between two particles $$2 mm$$ apart at any instant is

A
$${18}^{o}$$

B
$${36}^{o}$$

C
$${54}^{o}$$

D
$${72}^{o}$$

Solution

Given equation can be written as $$y = 6\sin(4\pi t- 0.2\pi x)$$
$$\therefore$$ Phase $$\phi = -0.2\pi x$$
$$\Rightarrow\phi_1 = -0.2\pi x_1$$ and $$\phi_2 = -0.2\pi x_2$$
$$\therefore$$ Phase difference $$\Delta\phi =|\phi_2 - \phi_1| =| -0.2\pi(x_2-x_1)|$$
given $$(x_2 - x_1) = 2mm$$ substituting in above equation we get $$\Delta\phi = 0.2\times{180^0}\times{2} = 72^0$$
Question 6
1 / -0

Two particles A and B execute simple harmonic motions of periods T and 5T / 4. They start from mean position. The phase difference between them when the particle A complete one oscillation will be:

A
$$0$$

B
$$\dfrac{\pi}{2}$$

C
$$\dfrac{2\pi}{5}$$

D
$$\dfrac{\pi}{6}$$

Solution

Equations of motion of the practicals are
$$X_1=A_1 sin \dfrac{2\pi}{T_1}t$$
and $$ X_2=A_2 sin \dfrac{25}{T_2}t$$
$$\therefore $$ Phase difference $$\Delta \phi =\left ( \dfrac{2\pi}{T_1}-\dfrac{2\pi}{T_2} \right )t$$
$$=\left ( \dfrac{2\pi}{T}-\dfrac{2\pi}{5T/4} \right )t$$
at t=T
$$\Delta \phi =\left ( 2 \pi -\dfrac{4 \times 2\pi}{5} \right )\frac{T}{T}=\dfrac{2\pi}{5}$$
Question 7
1 / -0

The phenomena during arising due to the superstition of waves is/are :

A
Beats

B
Stationary waves

C
Lissajous figures

D
All of these

Solution

Beat stationary waves, lissajous figures all correspond to the phenomena of superposition of waves.
Question 8
1 / -0

A progressive wave is represented as $$y=0.2\cos \pi (0.04 t+0.2x-\dfrac{\pi}{6})$$ where distance is expressed in cm and time in second. What will be the minimum distance between two particles having the phase difference of $$\dfrac{\pi}{2}$$?

A
4 cm

B
8 cm

C
25 cm

D
12.5 cm

Solution

On comparing the given equation with
$$y=a\cos (\omega t + kx - \phi)$$
we get, $$k=\dfrac{2\pi}{\lambda}=0.02$$
and $$\lambda = 100 cm$$
It is given that phase difference between particles $$\triangle \phi = \dfrac{\pi}{2}$$
So, the path difference between them,
$$\triangle p = \dfrac{\lambda}{2\pi}\times \triangle \phi$$
$$=\dfrac{\lambda}{2\pi}\times \dfrac{\pi}{2}=\dfrac{\lambda}{4}=\dfrac{100}{4}=25 cm$$
Question 9
1 / -0

With the propagation of a longitudinal wave through a material medium the quantities transmitted in the propagation direction are

A
Energy, momentum and mass

B
Energy

C
Energy and mass

D
Energy and linear momentum

Solution

$$\textbf{Explanation}$$
$$\bullet$$In longitudinal wave,particle vibrates in the direction of motion of wave.
$$\bullet$$When wave travels it means energy transfers from one end to other in the direction of propagation of wave.
$$\bullet$$Particles vibrate in the direction of propagation of wave so linear momentum also transmitted in the direction of propagation of wave.
Question 10
1 / -0

At a moment in a progressive wave, the phase of a particle executing SHM is $$\dfrac {\pi}{3}$$. Then the place of the particle $$15\ cm$$ ahead and at time $$\dfrac {T}{2}$$ will be, if the wavelength is $$60\ cm$$.

A
Zero

B
$$\dfrac {\pi}{2}$$

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

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

Solution

Let the phase of second particle is $$\phi$$. Hence the phase difference two particles is $$\triangle \phi = \dfrac {2\pi}{\lambda} \triangle x$$
$$\left (\phi - \dfrac {\pi}{3}\right ) = \dfrac {2\pi}{60}\times 15$$
$$\phi = \dfrac {\pi}{2} + \dfrac {\pi}{3}$$
$$\phi = \dfrac {5\pi}{6}$$.

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Re-Attempt Weekly Quiz Competition

Waves Test - 41

Result

Waves Test - 41

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SHARING IS CARING

Question 1

$$2\sqrt {2} m$$

$$4\sqrt {2}m$$

$$6\sqrt {2} m$$

$$8\sqrt {2}m$$

Solution

Question 2

$$\dfrac{\lambda}{3}$$

$$\dfrac{\lambda}{6}$$

$$\dfrac{\lambda}{9}$$

$$\lambda$$

Solution

Question 3

$$\dfrac{\pi}{2} cm$$

$$\dfrac{\pi}{4} cm$$

$$\dfrac{\pi}{8} cm$$

$$\dfrac{\pi}{16} cm$$

Solution

Question 4

$$200 cm$$

$$230 cm$$

$$210 cm$$

$$250 cm$$

Solution

Question 5

$${18}^{o}$$

$${36}^{o}$$

$${54}^{o}$$

$${72}^{o}$$

Solution

Question 6

$$0$$

$$\dfrac{\pi}{2}$$

$$\dfrac{2\pi}{5}$$

$$\dfrac{\pi}{6}$$

Solution

Question 7

Beats

Stationary waves

Lissajous figures

All of these

Solution

Question 8

4 cm

8 cm

25 cm

12.5 cm

Solution

Question 9

Energy, momentum and mass

Energy

Energy and mass

Energy and linear momentum

Solution

Question 10

Zero

$$\dfrac {\pi}{2}$$

$$\dfrac {5\pi}{6}$$

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

Solution

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