Sound Test

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Sound Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The separation between two consecutive crests in a transverse wave is $100 \ m$ . If wave velocity is $20 \ ms^{-1}$ , find the frequency of wave.

A
$0.1 \ Hz$

B
$0.2 \ Hz$

C
$0.3 \ Hz$

D
$0.5 \ Hz$

Solution

The distance between any two consecutive crests or trough of a wave is known as the wavelength of the wave.

Hence, here $\lambda = 100 \ m$
Also, given , $v=100 \ m/s$

The wavelength, wave velocity and frequency relationship is given as follows.

$\lambda =\dfrac { v }{ f }$ .

From this equation, frequency is derived as $f=\dfrac { v }{ \lambda }$ .

$f =\dfrac { 20\quad m/s }{ 100\quad m } = 0.2 \ Hz$ .
Question 2
1 / -0

State the approximate speed of ultrasound in air.

A
$320 \ m s^{-1}$

B
$330 \ m s^{-1}$

C
$340 \ m s^{-1}$

D
Can't say

Solution

The speed of ultrasound waves in a particular medium varies from 330 meters per second (m/sec) in air to 1450 m/sec in fat, 1570 m/sec in blood, and 4080 m/sec in the skull.
Hence, ultrasound travel through the air with speed of 330 m/s.
Question 3
1 / -0

Sound of frequency above $20kHz$ is called as

A
ultrasound

B
infrasound

C
breakable

D
unbreakable

Solution

Ultrasound is an oscillating sound pressure wave with a frequency greater than the upper limit of the human hearing range.
Ultrasound devices operate with frequencies from 20 kHz up to several gigahertz.
Hence, the ultrasonic waves are high frequency waves beyond human audibility.
Question 4
1 / -0

A sound of wave produces $40$ crests and $40$ troughs in $0.4s$ . What is the frequency of the wave?

A
$10\ Hz$

B
$50\ Hz$

C
$100\ Hz$

D
$150\ Hz$

Solution

One complete cycle consists of one crest and one trough.
The given source produces 40 crests and 40 troughs, hence it produces 40 wave cycles in the given interval 0.4 seconds.

Therefore, number of wave cycles produced in 1 second $=\quad \dfrac { 40\times 1 }{ 0.4 } \quad =\quad 100$

We know that the number of wave cycles produced in one second is equal to the frequency.

Hence, the frequency of the wave is 100 Hz.
Question 5
1 / -0

On the basis of following features identify correct option:
I. It is the characteristic of a sound.
II. It distinguishes an acute or a shrill note from a dull or flat note.

A
Loudness

B
Quality

C
Pitch

D
Both (A) and (B)

Solution

The sensation of a frequency is commonly referred to as the pitch of a sound. A high pitch sound corresponds to a high-frequency sound wave and a low pitch sound corresponds to a low-frequency sound wave.
Therefore, pitch is a characteristic of a sound wave that distinguishes an acute or a shrill note from a dull or flat note.
Question 6
1 / -0

Consider the following statement:
(a) Sound waves are longitudinal in nature.
(b) Sound waves cannot travel through vacuum.
(c) Sound waves are produced by oscillating charged particles only.
(d) Sound waves are electromagnetic waves.
Which of the following statements are correct ?

A
(a) and (b)

B
(b) and (c)

C
(a), (b) and (d)

D
(a),(b),(c) and (d)

Solution

Sound waves are longitudinal waves. It needs a medium to propagate. Sound waves are example of mechanical waves, not electromagnetic waves. Electromagnetic waves are created by the vibration of an electric charge.
Hence (a) and (b) are correct.
Question 7
1 / -0

A longitudinal wave is produced on a toy slinky. The wave travels at a speed of 30 cm/s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions of the slinky?

A
$0.015m$

B
$0.03m$

C
$0.045m$

D
$0.06m$

Solution

The relationship between the frequency, wavelength and speed of sound or velocity is given as follows.
$Wavelength\quad \lambda =\dfrac { wave\quad velocity\quad v }{ Frequency\quad \nu }$

In the question, the velocity of sound is 30 cm/s, that is 0.3 m/s and frequency=20 Hz is given.
The distance between two consecutive compression is equal to the wavelength of the wave, therefore,
$Wavelength\quad \lambda =\dfrac { v }{ \nu } =\dfrac { 0.3\quad m/s }{ 20\quad Hz } =0.015\quad m$ .
Hence, the wavelength of the sound wave is given as 0.015 m.
Question 8
1 / -0

A longitudinal wave travels at a speed of $0.3 \ ms^{-1}$ and the frequency of wave is $20 \ Hz$ . Find the separation between two consecutive compressions.

A
$0.5 \ cm$

B
$1 \ cm$

C
$1.5 \ cm$

D
$2 \ cm$

Solution

Given,
$v=0.3\ m/s\\ f=20 Hz$

The wavelength, wave velocity and frequency relationship are given as follows.
$\lambda =\dfrac { v }{ f }$ .

$=\dfrac { 0.3\quad m/s }{ 20\quad Hz } =0.015 m.$

That is $\quad 1.5\times { 10 }^{ -2 }m$

Hence, the separation between two consecutive compressions, i.e., the wavelength of the wave is 0.015 m or 1.5 cm
Question 9
1 / -0

A boat at anchor is rocked by waves whose consecutive crests are 100 m apart. The wave velocity of the moving crests is 20 m/s. What is the frequency of rocking of the boat?

A
$0.1\: Hz$

B
$0.2\: Hz$

C
$1\: Hz$

D
$2\: Hz$

Solution

The relationship between the frequency, wavelength and speed of sound or velocity is given as follows.
Frequency, $\nu =\cfrac { wave \: velocity, v }{ wavelength, \lambda }$
The distance between two consecutive crests is equal to the wavelength of the wave, therefore, $\lambda = 100 \:m$
So, $\nu = \cfrac{v}{\lambda}=\cfrac{20\:m/s}{100\:m}=0.2\:Hz$
Hence, the frequency of the wave is 0.2 Hz.
Question 10
1 / -0

A bat can hear sound at frequencies up to 120 kHz. Determine the wavelength of sound in the air at this frequency. Take the speed of sound in the air as 344 m/s :

A
$1.87\times 10^{-3}m$

B
$2.78\times 10^{-3}m$

C
$3.87\times 10^{-3}m$

D
$4.87\times 10^{-3}m$

Solution

The relationship between the frequency, wavelength and speed of sound or velocity is given as follows.
$$Frequency\quad \nu =\dfrac { wave\ velocity\quad v }{ wavelength\quad \lambda } $$
From the above equation, wavelength is derived as, $$Wavelength\quad \lambda =\dfrac { wave\quad velocity\ v }{ Frequency\quad \nu } $$

In the question, the velocity of sound is 344 m/s and frequency=120 kHz = $120\times { 10 }^{ 3 }Hz$ is given.
Therefore,
$Wavelength\quad \lambda =\dfrac { v }{ \nu } =\dfrac { 334\quad m/s }{ 120\times { 10 }^{ 3 }\quad Hz } =2.78\times { 10 }^{ -3 }m$ .

Hence, the wavelength of the sound wave is given as $2.78\times { 10 }^{ -3 }m$ .

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Sound Test - 34

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Sound Test - 34

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SHARING IS CARING

Question 1

0.1 Hz0.1 \ Hz0.1 Hz

0.2 Hz0.2 \ Hz0.2 Hz

0.3 Hz0.3 \ Hz0.3 Hz

0.5 Hz0.5 \ Hz0.5 Hz

Solution

Question 2

320 ms−1320 \ m s^{-1}320 ms−1

330 ms−1330 \ m s^{-1}330 ms−1

340 ms−1340 \ m s^{-1}340 ms−1

Can't say

Solution

Question 3

ultrasound

infrasound

breakable

unbreakable

Solution

Question 4

10 Hz10\ Hz10 Hz

50 Hz50\ Hz50 Hz

100 Hz100\ Hz100 Hz

150 Hz150\ Hz150 Hz

Solution

Question 5

Loudness

Quality

Pitch

Both (A) and (B)

Solution

Question 6

(a) and (b)

(b) and (c)

(a), (b) and (d)

(a),(b),(c) and (d)

Solution

Question 7

0.015m0.015m0.015m

0.03m0.03m0.03m

0.045m0.045m0.045m

0.06m0.06m0.06m

Solution

Question 8

0.5 cm0.5 \ cm0.5 cm

1 cm1 \ cm1 cm

1.5 cm1.5 \ cm1.5 cm

2 cm2 \ cm2 cm

Solution

Question 9

0.1 Hz0.1\: Hz0.1Hz

0.2 Hz0.2\: Hz0.2Hz

1 Hz1\: Hz1Hz

2 Hz2\: Hz2Hz

Solution

Question 10

1.87×10−3m1.87\times 10^{-3}m1.87×10−3m

2.78×10−3m2.78\times 10^{-3}m2.78×10−3m

3.87×10−3m3.87\times 10^{-3}m3.87×10−3m

4.87×10−3m4.87\times 10^{-3}m4.87×10−3m

Solution

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$0.1 \ Hz$

$0.2 \ Hz$

$0.3 \ Hz$

$0.5 \ Hz$

$320 \ m s^{-1}$

$330 \ m s^{-1}$

$340 \ m s^{-1}$

$10\ Hz$

$50\ Hz$

$100\ Hz$

$150\ Hz$

$0.015m$

$0.03m$

$0.045m$

$0.06m$

$0.5 \ cm$

$1 \ cm$

$1.5 \ cm$

$2 \ cm$

$0.1\: Hz$

$0.2\: Hz$

$1\: Hz$

$2\: Hz$

$1.87\times 10^{-3}m$

$2.78\times 10^{-3}m$

$3.87\times 10^{-3}m$

$4.87\times 10^{-3}m$