Waves — Page 2 | NEET Physics

Q11

Two cars moving in opposite directions approach each other with speed of 22 m s

-

1 and 16.5 m s

-

1 respectively. The driver of the first car blows a horn having a frequency 400 Hz. The frequency heard by the driver of the second car is (velocity of sound is 340 m s

-

1 )

A 361 Hz

B 411 Hz

C 448 Hz

D 350 Hz

Correct Answer

Option C

Solution

As we known from Doppler's Effect

{f_{apprent}} = {f_0}\left[ {{{v + {v_0}} \over {v - {v_s}}}} \right] = 400\left[ {{{340 + 16.5} \over {340 - 22}}} \right]

$\therefore$ f apprent = 448 Hz

Q12

The two nearest harmonics of a tube closed at one end and open at other end are 220 Hz and 260 Hz. What is the fundamental frequency of the system ?

A 20 Hz

B 30 Hz

C 40 Hz

D 10 Hz

Correct Answer

Option A

Solution

The difference of frequencies of closed pipe

{{2v} \over {4l}} = 260 - 220 = 40\,Hz

So, the fundamental frequency

f' = {v \over {4l}} = 20\,Hz

Q13

The second overtone of an open organ pipe has the same frquency as the first overtone of a closed pipe L metre long. The length of the open pipe will be

A L

B 2L

C

{L \over 2}

D 4L

Correct Answer

Option B

Solution

Second overtone of an open organ pipe (Third harmonic)

= 3 \times v{'_0} = 3 \times {v \over {2L'}}

First overtone of a closed organ pipe (Third harmonic)

= 3 \times {v_0} = 3 \times {v \over {4L}}

According to question,

3v{'_0} = 3{v_0} \Rightarrow 3 \times {v \over {2L'}} = 3 \times {v \over {4L}}

$\therefore$ L' = 2L

Q14

Three sound waves of equal amplitudes have frequencies (n

-

1), n, (n + 1). They superimpose to give beats. The number of beats produced per second will be

A 1

B 4

C 3

D 2

Correct Answer

Option D

Solution

(n – 1) and (n + 1) suppose to form frequency n n and n will be at resonance n – 1 and n produce 1 beat n + 1 and n produce 1 beat $\therefore$ Number of beats formed are ‘2’

Q15

A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 m s

-

1 . Then, the frequency of sound that the observer hears in the echo reflected from the cliff is (Take velocity of sound in air = 330 m s

-

1 )

A 838 Hz

B 885 Hz

C 765 Hz

D 800 Hz

Correct Answer

Option A

Solution

According to Doppler's effect in sound Apparent frequency,

n' = {v \over {v - {v_S}}}{n_0}

= {{330} \over {330 - 15}}\left( {800} \right) = {{330 \times 800} \over {315}} = 838\,Hz

The frequency of sound observer hears in the echo reflected from the cliff is 838 Hz.

Q16

An air column, closed at one end open at the other, resonates with a tuning fork when the smallest length of the column is 50 cm. The next larger length of the column resonating with the same tuning fork is

A 150 cm

B 200 cm

C 66.7 cm

D 100 cm

Correct Answer

Option A

Solution

From figure, First harmonic is obtained at

l = {\lambda \over 4} = 50\,cm

Third harmonic is obtained for resonance,

l' = {{3\lambda } \over 4} = 3 \times 50 = 150\,cm

Q17

A uniform rope of length L and mass m 1 hangs vertically from a rigid support. A block of mass m 2 is attached to the free end of the rope. A transverse pulse of wavelength

\lambda

1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is

\lambda

2 . The ratio

\lambda

2 /

\lambda

1 is

A

\sqrt {{{{m_2}} \over {{m_1}}}}

B

\sqrt {{{{m_1} + {m_2}} \over {{m_1}}}}

C

\sqrt {{{{m_1}} \over {{m_2}}}}

D

\sqrt {{{{m_1} + {m_2}} \over {{m_2}}}}

Correct Answer

Option D

Solution

Wavelength of pulse at the lower end

{\lambda _1} \propto velocity({v_1}) = \sqrt {{{{T_1}} \over \mu }}

Similarly,

{\lambda _1} \propto {v_2}

= \sqrt {{{{T_2}} \over \mu }}

$\therefore$

{{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{T_2}} \over {{T_1}}}} = \sqrt {{{\left( {{m_1} + {m_2}} \right)g} \over {{m_2}g}}}

= \sqrt {{{{m_1} + {m_2}} \over {{m_2}}}}

Q18

A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is

A 10.5 Hz

B 105 Hz

C 155 Hz

D 205 Hz

Correct Answer

Option B

Solution

In a stretched string all multiples of frequencies can be obtained i.e., if fundamental frequency is n then higher frequencies will be 2n, 3n, 4n ...

So, the difference between any two successive frequencies will be 'n' According to question, n = 420 – 315 = 105Hz So the lowest frequency of the string is 105 Hz.

Q19

The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is

A 120 cm

B 140 cm

C 80 cm

D 100 cm

Correct Answer

Option A

Solution

Fundamental frequency of closed organ pipe

{V_c} = {V \over {4{l_c}}}

Fundamental frequency of open organ pipe

{V_0} = {V \over {2{l_0}}}

Second overtone frequency of open organ pipe

= {{3V} \over {2{l_0}}}

{V \over {4{l_c}}} = {{3V} \over {2{l_0}}}

\Rightarrow {l_0} = 6{l_c} = 6 \times 20 = 120

cm

Q20

A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m s

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1 at an angle of 60 o with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 m s

-

1 ), is

A 106 Hz

B 97 Hz

C 100 Hz

D 103 Hz

Correct Answer

Option D

Solution

Here, original frequency of sound, f 0 = 100 Hz Speed of source V s = 19.4 cos 60° = 9.7 From Doppler's formula

{f^1} = {f_0}\left( {{{V - {V_0}} \over {V - {V_S}}}} \right)

{f^1} = 100\left( {{{V - 0} \over {V - \left( { + 9.7} \right)}}} \right)

{f^1} = 100{V \over {V\left( {1 - {{9.7} \over V}} \right)}} = {{100} \over {\left( {1 - {{9.7} \over {330}}} \right)}}

= 103Hz Apparent frequency f 1 = 103 Hz