Waves Test

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Waves Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

An ultrasonic scanner is used in a hospital to detect tumours in tissue. The working frequency of the scanner is 4.2 mega Hz. The velocity of sound in the tissue is $$1.7 kms^{-1}$$. The wavelength of sound in the tissue is nearest to

A
$$\displaystyle 4 \times 10^{-3}$$ m

B
$$\displaystyle 8 \times 10^{-3}$$ m

C
$$\displaystyle 4 \times 10^{-4}$$ m

D
$$\displaystyle 8 \times 10^{-4}$$ m

Solution

Let,

The working frequency of the scanner is $$\nu = 4.2 MHz = 4.2 \times 10^6 Hz$$.

The velocity of sound in the tissue is $$v = 1.7 kms^{-1} = 1.7 \times 10^3 ms^{-1}$$

the wavelength of sound is given by

$$\lambda = \dfrac{v}{\nu}$$

$$\lambda = \dfrac{1.7 \times 10^3}{4.2 \times 10^6}$$

$$\lambda = 0.404 \times 10^{-3}$$

$$\lambda \approx 4 \times 10^{-4} m$$
Question 2
1 / -0

In a Kundt's tube experiment, the heaps of lycopodium powder are collected at $$20\ cm$$ separations. The frequency of tuning fork used is

A
$$660\ Hz$$

B
$$825\ Hz$$

C
$$775\ Hz$$

D
$$915\ Hz$$

Solution

The in kundt's tube powder is collected where nodes are formed. The separation between nodes is given as

$$d=\dfrac{\lambda}{2} =20 cm$$

$$\lambda=40 cm$$

$$f=\dfrac{c}{\lambda}$$$$=\dfrac{330}{.4} =825 \ Hz$$
Question 3
1 / -0

Two wires with different densities are joined at $$x = 0$$. An incident wave $$Y_i = A_i\sin(\omega t -k_1x)$$ travelling from left to right is partially reflected and partially transmitted $$(Y_t = A_t\sin(\omega t -k_2x))$$ at $$x = 0$$. If the amplitude of transmitted wave is $$A_t$$ then $$^{\Large A_t}/_{\Large A_i}$$ is

A
$$\displaystyle\frac{2k_1}{k_1+k_2}$$

B
$$\displaystyle\frac{2k_2}{k_1+k_2}$$

C
$$\displaystyle\frac{k_1-k_2}{k_1+k_2}$$

D
$$\displaystyle\frac{\sqrt{k_1}-\sqrt{k_2}}{\sqrt{k_1}+\sqrt{k_2}}$$

Solution

Let the equation of reflected wave be $$A_rsin(\omega t+k_1x)$$
The sign is reversed to include the reversed direction of the reflected wave.
From continuity of the wave at the interface $$x=0$$, we get,
$$A_i+A_r=A_t$$
Also the first derivative of the wave are continuous at the boundary, hence,
$$k_1(A_i-A_r)=k_2A_t$$
$$\implies A_i-A_r=\dfrac{k_2}{k_1}A_t$$
Rearranging the two equations gives,
$$\dfrac{A_t}{A_i}=\dfrac{2k_1}{k_1+k_2}$$
Question 4
1 / -0

The displacement $$y$$ in centimeters is given in terms of time $$t$$ in second by the equation: $$y=3\sin 3.14t+4\cos 3.14 t$$, then the amplitude of SHM is

A
3 cm

B
4 cm

C
5 cm

D
7 cm

Solution

When the displacement of a SHM is $$y=a\sin wt+b\cos wt$$, the amplitude of the SHM will be $$A=\sqrt{a^2+b^2}$$
Here, $$a=3, b=4$$ so amplitude $$A=\sqrt{3^2+4^2}=5 $$ cm
Question 5
1 / -0

The equation of displacement of a harmonic oscillator is $$x=3\sin{\omega t}+4\cos{\omega t}$$. The amplitude of the particles will be

A
1

B
5

C
7

D
12

Solution

In the equation of superposition of SHMs or Harmonic oscillator, $$x=asin\omega t+bsin\omega t$$, the resultant amplitude is given by $$A=\sqrt{a^{2}+b^{2}}$$. Substituting a=3 and b=4, we get final answer as Option B, i.e. 5
Question 6
1 / -0

The wavelength of a sound wave is reduced by 50%. Then the percentage change in its frequency will be

A
100%

B
200%

C
400%

D
800%

Solution

The wavelength of sound wave is
$$\lambda = \dfrac{v}{\nu}$$
$$\nu = \dfrac{v}{\lambda}$$
where, v is the velocity of sound waves and $$\nu$$ is the frequency of sound waves.
When the wavelength of a sound wave is reduced by 50%
$$\lambda - \dfrac{50}{100}\lambda = \dfrac{v}{\nu'}$$
$$\dfrac{1}{2}\lambda = \dfrac{v}{\nu'}$$
$$\nu' = \dfrac{2v}{\lambda}$$
The percent change in frequency is
$$\dfrac{\nu - \nu'}{\nu} \times 100 = \dfrac{\dfrac{2v}{\lambda} - \dfrac{v}{\lambda}}{\dfrac{v}{\lambda}} \times 100$$
$$\Rightarrow (2 - 1) \times 100 = 100$$%
Question 7
1 / -0

If the frequency of a sound wave is increased by 25%, then the change in its wavelength will be

A
25% decrease

B
20% decrease

C
20% increase

D
25% increase

Solution

The frequency of wave is given by

$$\nu = \dfrac{v}{\lambda}$$

$$\lambda = \dfrac{v}{\nu}$$

When the frequency of a sound wave is increased by 25%, then the new
wavelength is

$$\lambda' = \dfrac{v}{\nu+\dfrac{25}{100}\nu}$$

$$\lambda' = \dfrac{v}{\nu+\dfrac{1}{4}\nu}$$

$$\lambda' = \dfrac{v}{\dfrac{5}{4}\nu}$$

$$\lambda' = \dfrac{4v}{5\nu}$$

Hence, the percent change in wavelength is

$$\dfrac{\lambda - \lambda'}{\lambda} \times 100 = \dfrac{\dfrac{v}{\nu} - \dfrac{4v}{5\nu}}{\dfrac{v}{\nu}} \times 100$$

$$\Rightarrow -\dfrac{1}{4} \times 100 = -20$$%.

Hence, wavelength decreases by 20%
Question 8
1 / -0

A sound wave of frequency $$500\ \text{Hz}$$ covers a distance of $$1000\ \text{m}$$ in $$5\ \text{s}$$ between points $$\displaystyle x$$ and $$\displaystyle y$$. Then the number of waves between $$\displaystyle x$$ and $$\displaystyle y$$ are

A
5000

B
2500

C
100

D
500

Solution

$$\nu= 500\ \text{Hz}$$

$$x= 1000$$

$$t= 5\ \text{s}$$

velocity of sound $$v=\dfrac{x}{t} = 200\ \text{m/s}$$

But $$v= \nu \lambda$$

Hence, $$\lambda = \dfrac{2}{5}\ \text{m/s}$$

Now $$n\lambda = x\ \text{or}\ n = 2500$$

Ans: B
Question 9
1 / -0

A sonometer wire, $$100\ \text{cm}$$ in length has a fundamental frequency of $$330\ \text{Hz}$$. The velocity of propagation of transverse waves along this wire is :

A
$$330\ \text{ms}^{-1}$$

B
$$660\ \text{ms}^{-1}$$

C
$$115\ \text{ms}^{-1}$$

D
$$990\ \text{ms}^{-1}$$

Solution

For the fundamental frequency $$(f_0=330 Hz),\ \lambda=2 L=200\ \text{cm} = 2\ \text{m}$$

then from the formula, $$v=f_0 \times \lambda= 330 \times 2.0= 660 \text{ms}^{-1}$$
Question 10
1 / -0

Interference event is observed

A
only in transverse waves

B
only in longitudinal waves

C
in both types of waves

D
none

Solution

Interference occurs for both transverse and longitudinal waves.
Interference does not indicate whether light is transverse or longitudinal since interference occurs for both types of waves.

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Waves Test - 29

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Waves Test - 29

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SHARING IS CARING

Question 1

$$\displaystyle 4 \times 10^{-3}$$ m

$$\displaystyle 8 \times 10^{-3}$$ m

$$\displaystyle 4 \times 10^{-4}$$ m

$$\displaystyle 8 \times 10^{-4}$$ m

Solution

Question 2

$$660\ Hz$$

$$825\ Hz$$

$$775\ Hz$$

$$915\ Hz$$

Solution

Question 3

$$\displaystyle\frac{2k_1}{k_1+k_2}$$

$$\displaystyle\frac{2k_2}{k_1+k_2}$$

$$\displaystyle\frac{k_1-k_2}{k_1+k_2}$$

$$\displaystyle\frac{\sqrt{k_1}-\sqrt{k_2}}{\sqrt{k_1}+\sqrt{k_2}}$$

Solution

Question 4

3 cm

4 cm

5 cm

7 cm

Solution

Question 5

1

5

7

12

Solution

Question 6

100%

200%

400%

800%

Solution

Question 7

25% decrease

20% decrease

20% increase

25% increase

Solution

Question 8

5000

2500

100

500

Solution

Question 9

$$330\ \text{ms}^{-1}$$

$$660\ \text{ms}^{-1}$$

$$115\ \text{ms}^{-1}$$

$$990\ \text{ms}^{-1}$$

Solution

Question 10

only in transverse waves

only in longitudinal waves

in both types of waves

none

Solution

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