Waves Test

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Waves Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

Two pulses in a stretched string whose centres are initially 8 cm apart are moving towards each other as shown in Fig. 7.80. The speed of each pulse is 2 cm/s. After 2 s the total energy of the pulses will be

A
zero

B
purely kinetic

C
purely potential

D
partly kinetic and partly potential

Solution

b. After 2 s, the two pules will nullify each other. As string now becomes straight, there will be no deformation in the string. In such a situation, the string will not have potential energy at any point. The whole energy will be kinetic.
Question 2
1 / -0

n waves are produced on a string in 1 s. When the radius of the string is doubled and the tension is maintained the same, the number of waves produced in 1 s for the same harmonic will be

A
2n

B
$$\frac {n} {3}$$

C
$$\frac {n} {2}$$

D
$$\frac {n} {\sqrt{2}}$$

Solution

d. Given that the frequency of wave produced in the string is l/n.
$$ \therefore \dfrac{1} {n} = \dfrac{1} {2 \pi} \sqrt {\dfrac{T} {m}}$$
Now T = 2T

Therefore, new frequency is
$$ f = \dfrac {1} {2\pi} \sqrt {\dfrac {2T} {m}} = \sqrt {2} \times \dfrac{1} {n}$$

Therefore, the number of waves produced per second is
$$ \dfrac {1} {f} = \dfrac {n} {\sqrt{2}}$$
Question 3
1 / -0

One end of a 2.4 m string is held fixed and the other end is attached to a weightless ring that can slide along a frictionless rod as shown in Fig. 7.86. The three longest possible wavelength for standing waves in this string are respectively

A
4.8 m, 1.6 m and 0.96 m

B
9.6 m, 3.2 m and 1.92 m

C
2.4 m, 0.8 m and 0.48 m

D
1.2 m, 0.4 m and 0.24 m

Solution

b. When the end of the string is free to move, the string being attached to weightless ring that can slide freely along the rod, the phase of reflected pulse is unchanged antinode is formed at the ring.
$$ l = \frac{\lambda} {4} \Rightarrow \lambda_1 = 4l = 9.6 \space m$$
$$\lambda_2 = \frac {4l} {3} = 3.2 \space m$$
$$\lambda_3 = \frac{4l} {5} = \frac {9.6} {5} = 1.92 \space m$$
Question 4
1 / -0

Which of the following travelling wave will produce standing wave, with nodes at x = 0, when superimposed on $$y = A \sin{(\omega t - kx)}$$

A
$$A \sin{(\omega t + kx)}$$

B
$$A \sin{(\omega t + kx + \pi)}$$

C
$$A \cos{(\omega t + kx)}$$

D
$$A \cos{(\omega t + kx + \pi)}$$

Solution

b. Substituting x = 0, we have given wave $$y = A \sin{\omega t }$$ at x = 0 other should have $$y = - A \sin{\omega t }=A\sin(\omega +\pi)$$ equation so displacement may be zero at all the time. Hence, option (b) is correct.
Question 5
1 / -0

Let the two waves $$y_1 = A \sin {(kx - \omega t)}$$ and $$y_2 = A \sin {(kx + \omega t)}$$ form a standing wave on a string. Now if an additional phase difference of $$\phi$$ is created between two waves, then

A
the standing wave will have a different frequency

B
the standing wave will have a different amplitude for a given point

C
the spacing between two consecutive nodes will change

D
none of the above

Solution

b. Initially the standing wave equation is
$$y_1 = 2A \sin {kx} \cos {\omega t}$$
If phase difference $$\phi$$ is added to one of waves, then resulting standing wave equation is
$$y = 2A \sin {\left (kx + \frac{\phi} {2} \right )} \cos {\left (\omega t - \frac{\phi} {2} \right )}$$
Here, frequency does not change and also spacing between two successive nodes does not change as its value for both is $$\pi/k$$. But for a particle, in standing wave, amplitude changes.
Question 6
1 / -0

Microwaves from a transmitter are directed normally towards a plane reflector. A detector moves along the normal to the reflector. Between positions of 14 successive maxima, the detector travels a distance 0.14 m. If the velocity of light is $$ 3 \times 10^{8}$$ m/s, find the frequency of the transmitter.

A
$$ 1.5 \times 10^{10} Hz$$

B
$$ 10^{10} Hz$$

C
$$ 3 \times 10^{10} Hz$$

D
$$ 6 \times 10^{10} Hz$$

Solution

a. If detector moves x distance, distance from direct sound increases by x and distance from reflected sound decreases by x so path difference created = 2x
$$ 2(0.14) = 14 \lambda = 14 \space c/f$$
$$ f = \frac {14 \times 3 \times 10^{8}} {0.14 \times 2} = 1.5 \times 10^{10}Hz$$
Question 7
1 / -0

Which of the following are transferred from one place to another place by the waves ?

A
mass

B
wavelength

C
velocity

D
energy

Solution

Energy is transferred from one place to another by the propagation of waves.
Question 8
1 / -0

Two waves are given by $$y_1 = a \sin \left (\omega t - kx \right )$$ and $$y_2 = a \cos \left (\omega t - kx \right )$$. The phase difference between the two waves is

A
$$\frac{\pi} {4}$$

B
$$\pi$$

C
$$\frac{\pi} {8}$$

D
$$\frac{\pi} {2}$$

Solution

The two equations of the waves are:
$$y_1 = a \sin \left (\omega t - kx \right )$$
and $$y_2 = a \cos \left (\omega t - kx \right )$$

The second equation can be represented as:
$$= a \sin \left (\omega t - kx + \dfrac{\pi} {2} \right )$$

Hence phase difference between these two is $$\dfrac{\pi} {2}$$
Question 9
1 / -0

If amplitude of waves at distance r from a point source is A, the amplitude at a distance 2r will be

A
$$2A$$

B
$$A$$

C
$$A/2$$

D
$$A/4$$

Solution

We know that the relation between the amplitude and the distance is given as:
$$I \propto a^{2} \propto \dfrac{1} {d^{2}} $$

Therefore, the amplitiude at a distance of $$2r$$ will be:
$$\Rightarrow a \propto \dfrac{1} {d}$$

$$\dfrac{A_1}{A_2}=\dfrac{r_2}{r_1}$$

$$A_2=\dfrac{r}{2r}$$

$$A_2=\dfrac{A}{2}$$
Question 10
1 / -0

Two waves are propagating to the point $$P$$ along a straight line produced by two sources $$A$$ and $$B$$ of simple harmonic and of equal frequency. The amplitude of every wave at $$P$$ is $$a$$ and the phase of A is ahead by $$\dfrac{pi}{3}$$ than that of $$B$$ and the distance $$AP$$ is greater than $$BP$$ by 50 $$cm$$. Then the resultant amplitude at the point $$P$$ will be, if the wavelength is 1 meter

[BVP 2003]

A
$$2a$$

B
$$\sqrt{3}$$

C
$$a\sqrt{2}$$

D
$$a$$

Solution

Path difference $$(\triangle x) = 50 cm = \dfrac{1}{2}m$$
$$\therefore$$ Phase difference $$\triangle \phi = \dfrac{2 \pi}{\lambda}\triangle x \Rightarrow \phi = \dfrac{2pi}{1} \dfrac{1}{2} = \pi $$
Total phase difference $$= \pi - \dfrac{\pi}{3} = \dfrac{2 \pi}{3}$$
$$\Rightarrow A = \sqrt{a^2 + a^2 + 2a^2 cos(2 \pi / 3)} = a$$

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Waves Test - 68

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Waves Test - 68

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SHARING IS CARING

Question 1

zero

purely kinetic

purely potential

partly kinetic and partly potential

Solution

Question 2

2n

$$\frac {n} {3}$$

$$\frac {n} {2}$$

$$\frac {n} {\sqrt{2}}$$

Solution

Question 3

4.8 m, 1.6 m and 0.96 m

9.6 m, 3.2 m and 1.92 m

2.4 m, 0.8 m and 0.48 m

1.2 m, 0.4 m and 0.24 m

Solution

Question 4

$$A \sin{(\omega t + kx)}$$

$$A \sin{(\omega t + kx + \pi)}$$

$$A \cos{(\omega t + kx)}$$

$$A \cos{(\omega t + kx + \pi)}$$

Solution

Question 5

the standing wave will have a different frequency

the standing wave will have a different amplitude for a given point

the spacing between two consecutive nodes will change

none of the above

Solution

Question 6

$$ 1.5 \times 10^{10} Hz$$

$$ 10^{10} Hz$$

$$ 3 \times 10^{10} Hz$$

$$ 6 \times 10^{10} Hz$$

Solution

Question 7

mass

wavelength

velocity

energy

Solution

Question 8

$$\frac{\pi} {4}$$

$$\pi$$

$$\frac{\pi} {8}$$

$$\frac{\pi} {2}$$

Solution

Question 9

$$2A$$

$$A$$

$$A/2$$

$$A/4$$

Solution

Question 10

$$2a$$

$$\sqrt{3}$$

$$a\sqrt{2}$$

$$a$$

Solution

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