Waves Test

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Waves Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Two wires made of the same material, one thick and 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 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 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 decreases.
Question 2
1 / -0

Spherical wavefronts, emanating from a point source, Strike a plane reflecting surface. What will happen to these wave fronts, immediately after reflection ?

A
They will remain spherical with the same curvature, both in magnitude and sign.

B
They will become plane wave fronts.

C
They will remain spherical, with the same curvature, but sign of curvature reversed.

D
They will remain spherical, but with different curvature, both in magnitude and sign.

Solution

The wavefront will remain to be spherical after reflection from the surface but their direction of propagation will be revered.
Question 3
1 / -0

Which of the following wave function does represent a travelling wave?

A
$$y=$$tan$$(x - vt)^2$$

B
$$y=$$log$$(x- vt)$$

C
$$y=\dfrac{1}{x+vt}$$

D
None of these

Solution

Travelling wave is the temporary wave that creates a disturbance and moves along the transmission line at a constant speed. These waves occur for a short duration, but cause a much disturbance in the line. The travelling waves can be represented mathematically in many ways.

The basic requirements for a wave function to represent a travelling wave is that for all values of x and t, wave function must have a finite value. Out of the given functions for y, no one satisfies this condition ( i.e. tan, log and inverse). Therefore, none can represent a travelling wave.

Thus the correct option is D.
Question 4
1 / -0

A vibratory motion is represented by
$$ x = 2A cos \omega t + A cos [\omega t + \frac{\pi}{2}] + A cos (\omega t + \pi) + \frac{A}{2} cos [\omega t + \frac{3 \pi}{2}]$$.
The resultant amplitude of the motion is

A
$$\dfrac{9A}{2}$$

B
$$\dfrac{\sqrt5A}{2}$$

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

D
$$2A$$

Solution

We can solve this by a vector, See the figure I have given.
Resultant, see the figure ii.
$$A_{net}=\sqrt{\left(2A-A\right)^2+{\left(A-\cfrac{A}{2}\right)^2}}=\sqrt{A^2+\cfrac{A^2}{4}}=\cfrac{\sqrt{5}A}{2}$$
Question 5
1 / -0

A particle executes SHM of period $$12$$s. After two seconds, it passes through the centre of oscillation, the velocity is found to be $$3.142 cm s^{-1}$$. The amplitude of oscillation is:

A
$$6$$cm

B
$$3$$ cm

C
$$24$$cm

D
$$12$$cm

Solution

Let $$y = A sin \omega t$$

Then $$v = \dfrac{dy}{dt} = \omega A cos \omega t = \dfrac{2 \pi}{T} A cos \dfrac{2 \pi}{T}t$$

$$\therefore 3.142 = \dfrac{2 \times 3.142}{12} \times A cos \dfrac{2 \pi}{12} \times 2$$

$$\therefore A = 12cm$$
Question 6
1 / -0

The function $$ sin \omega t - cos \omega t$$ represents

A
a simple harmonic motion with a period $$\frac{\pi}{\omega}$$.

B
a simple harmonic motion with a period $$\frac{2\pi}{\omega}$$.

C
a periodic, but not simple harmonic motion with a period $$\frac{\pi}{\omega}$$.

D
a periodic, but not simple harmonic motion with a period $$\frac 2{\pi}{\omega}$$.

Solution

$$x=\sin \omega t-\cos \omega t$$
$$\Longrightarrow x=2\cos \omega t=2\sin (\omega t+\cfrac{\pi}{2})\quad \quad \quad \quad [\sin C-\sin D=2\cos (\cfrac{C+D}{2})\sin (\cfrac{C-D}{2})]$$
On comparing this with equation of $$SHM:$$ $$x=A\sin (\omega t+\phi)$$
$$\Longrightarrow A=2\quad\quad \omega \text{(Angular frequency)}$$
$$\Longrightarrow T=\cfrac{2\pi}{\omega}$$
Question 7
1 / -0

there are $$26$$ tuning forks arranged in the decreasing order of their frequencies. Each tuning fork gives $$3$$ beats with the next. The first one is octave of the last. What is the frequency of $$18^{th}$$ tuning fork?

A
$$100 Hz$$

B
$$99 Hz$$

C
$$96 Hz$$

D
$$103 Hz$$

Solution

Let the frequency of first tuning fork is $$\upsilon$$. The frequencies of other tuning forks are
$$(\upsilon - 3), (\upsilon - 2 \times3), ......,(\upsilon - 17 \times 3), .....,(\upsilon - 25 \times 3)$$.
As per given condition,
$$\upsilon = 2(\upsilon - 25 \times 3)$$ or $$\upsilon = 2\upsilon - 25 \times 6$$
or $$\upsilon = 25 \times 6 = 150 Hz$$
The frequency of the $$18^{th}$$ tuning fork
$$\upsilon - 17 \times = 150 - 51 = 99 Hz$$
Question 8
1 / -0

In a certain oscillatory system, the amplitude of motion is $$5$$m and the time period is $$4$$s. The time taken by the particle for passing between points which are at distance of $$4$$m and $$2$$m from the centre and on the same side of it will be

A
$$0.30$$s

B
$$0.32$$

C
$$0.33$$

D
$$0.35$$

Solution

Let the motion be represented by
$$x=A\cos (wt)$$
Given amplitude $$(A)=5$$
$$w=\dfrac{2\pi}{7}$$ with periodic time $$T=45$$
$$\therefore w=\dfrac{2\pi}{4}=\dfrac{\pi}{2}$$
Also $$x_{1}=4\ cm$$
$$x_{2}=2\ m$$
Let $$t=t_{4}$$ for displacement of $$4\ m$$
$$\therefore 4=5\cos \left(\dfrac{\pi}{2}\times t_{4}\right)$$
$$\therefore 0.8=\cos \left(\dfrac{\pi}{2}t_{4}\right)$$ $$\therefore t_{4}=\dfrac{36.83}{90}=$$
For displacement of $$2\ m$$ let $$t=t_{2}$$
$$\therefore 2=5\cos\left(\dfrac{\pi}{2}t_{2}\right)$$
$$\therefore 0.4\cos\left(\dfrac{\pi t_{2}}{2}\right)$$ $$\therefore t_{2}=\dfrac{66.5}{90}=$$
$$\therefore$$ Time to travel from $$4\ m$$ to $$2\ m$$ is
$$=t_{4}-t_{2}$$
$$=\dfrac{29.67}{90}=0.3248$$ seconds
$$\approx 0.33$$ seconds
Question 9
1 / -0

The equation of a wave is given by
$$y= 10$$ sin$$\left(\dfrac{2\pi}{45} t + \alpha \right)$$.
If the displacement is $$5 cm$$ at $$t=0$$, the the total phase at $$t=7.5 s$$ is

A
$$\dfrac{\pi}{3}$$

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

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

D
$$\pi$$

Solution

The given equation of wave is
$$y= 10$$ sin $$\left(\dfrac{2\pi}{45} t + \alpha \right)$$

At $$t= 0, y= 5 cm$$

$$\therefore 5= 10$$ sin$$\alpha $$

$$\dfrac{1}{2}=$$ sin $$\alpha$$ or sin $$\left(\dfrac{\pi}{6}\right)=$$ sin$$\alpha$$

$$\alpha = \dfrac{\pi}{6}$$ ......(i)

hence, the total phase at $$t$$ = $$7.5 s=\left(\dfrac{15}{2}s\right)$$ is

$$\phi=\dfrac{2\pi}{45} \times \dfrac{15}{2} + \alpha =\dfrac{\pi}{3} + \dfrac{\pi}{6}$$ (Using (i))

$$\alpha =\dfrac{3\pi}{6} = \dfrac{\pi}{2}$$
Question 10
1 / -0

The displacement of a particle executing simple harmonic motion is given by
$$x = 3 sin [2 \pi t + \frac{\pi}{4}]$$ where x is in metres and t is in seconds. The amplitude and maximum speed of the particle is:

A
$$3m, 2 \pi m s^{-1}$$

B
$$3m, 4 \pi m s^{-1}$$

C
$$3m, 6 \pi m s^{-1}$$

D
$$3m, 8 \pi m s^{-1}$$

Solution

The given equation of SHM is

$$x = 3sin[2 \pi t + \dfrac{\pi}{4}]$$

Compare the given equation with standard equation of SHM
$$x = A sin(\omega t + \phi)$$

we get, $$A = 3m, \omega = 2 \pi s^{-1}$$

$$\therefore Maximum \,\, speed, v_{max} = A \omega = 3 m \times 2 \pi s^{-1} = 6 \pi m s^{-1}$$

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Waves Test - 44

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Waves Test - 44

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SHARING IS CARING

Question 1

Frequency only

Speed of propagation only

Wavelength only

The speed of propagation as well as the wavelength

Solution

Question 2

They will remain spherical with the same curvature, both in magnitude and sign.

They will become plane wave fronts.

They will remain spherical, with the same curvature, but sign of curvature reversed.

They will remain spherical, but with different curvature, both in magnitude and sign.

Solution

Question 3

$$y=$$tan$$(x - vt)^2$$

$$y=$$log$$(x- vt)$$

$$y=\dfrac{1}{x+vt}$$

None of these

Solution

Question 4

$$\dfrac{9A}{2}$$

$$\dfrac{\sqrt5A}{2}$$

$$\dfrac{5}{2}A$$

$$2A$$

Solution

Question 5

$$6$$cm

$$3$$ cm

$$24$$cm

$$12$$cm

Solution

Question 6

a simple harmonic motion with a period $$\frac{\pi}{\omega}$$.

a simple harmonic motion with a period $$\frac{2\pi}{\omega}$$.

a periodic, but not simple harmonic motion with a period $$\frac{\pi}{\omega}$$.

a periodic, but not simple harmonic motion with a period $$\frac 2{\pi}{\omega}$$.

Solution

Question 7

$$100 Hz$$

$$99 Hz$$

$$96 Hz$$

$$103 Hz$$

Solution

Question 8

$$0.30$$s

$$0.32$$

$$0.33$$

$$0.35$$

Solution

Question 9

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

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

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

$$\pi$$

Solution

Question 10

$$3m, 2 \pi m s^{-1}$$

$$3m, 4 \pi m s^{-1}$$

$$3m, 6 \pi m s^{-1}$$

$$3m, 8 \pi m s^{-1}$$

Solution

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