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WAVES

172978 A car, sounding a horn of frequency $1000 \mathrm{~Hz}$, is moving directly towards a huge wall at a speed of $15 \mathrm{~m} / \mathrm{s}$. If the speed of sound is $340 \mathrm{~m} / \mathrm{s}$, then the frequency of echo heard by the driver is

1 $1046 \mathrm{~Hz}$

2 $954 \mathrm{~Hz}$

3 $1092 \mathrm{~Hz}$

4 $908 \mathrm{~Hz}$

CG PET -2018

WAVES

172973 A source of sound $S$ in the form of a block kept on a smooth horizontal surface is connected to a spring, as shown in the figure. If the spring oscillates with an amplitude of $50 \mathrm{~cm}$ along horizontal between the wall and the observer $o$, the maximum frequency heard by the observer is $12.5 \%$ more, than the minimum frequency heard by him. If the mass of the source of sound is $100 \mathrm{~g}$, the force constant of the spring is
(Speed of sound in air is $340 \mathrm{~ms}^{-1}$ )

1 $40 \mathrm{Nm}^{-1}$

2 $80 \mathrm{Nm}^{-1}$

3 $160 \mathrm{Nm}^{-1}$

4 $320 \mathrm{Nm}^{-1}$

Explanation:

C Given data,
Amplitude of spring $(A)=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$
Maximum frequency heard by the observer is $12.5 \%$ more than the minimum frequency
$\qquad \mathrm{n}_{\max }=1.125 \mathrm{n}_{\text {min }}$
$\text { Mass of source }(\mathrm{m})=100 \mathrm{~g}=\frac{100}{1000}=0.10 \mathrm{~kg}$
Speed of sound $\mathrm{v}=340 \mathrm{~m} / \mathrm{s}$

Apparent frequency $(\mathrm{n})=\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}$
For $\quad \mathrm{n}_{\max }$
$\text { For } \quad \mathrm{v}_{\mathrm{s}}=\mathrm{v}_{\mathrm{s} \max }$
$\mathrm{n}_{\min },$
$\mathrm{v}_{\mathrm{s}}=-\mathrm{v}_{\mathrm{s} \max }$
$\mathrm{n}_{\max }=\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}$
$\mathrm{n}_{\text {min }}=\frac{\eta_{\mathrm{o}} \mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }}$
On dividing Equation (i) \& (ii), we get-
$\frac{\mathrm{n}_{\text {max }}}{\mathrm{n}_{\text {min }}}=\frac{\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}}{\frac{\mathrm{n}_{\mathrm{o}} \mathrm{V}}{\mathrm{v}+\mathrm{v}_{\mathrm{s} \text { max }}}}$
$\frac{1.125 \mathrm{n}_{\text {min }}}{\mathrm{n}_{\text {min }}}=\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}$
$1.125\left(\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }\right)=\left(\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }\right)$
$0.125 \mathrm{v}=2.125 \mathrm{v}_{\mathrm{s} \max }$
$\mathrm{v}_{\mathrm{s} \max }=\frac{0.125 \times 340}{2.125}$
$\mathrm{v}_{\mathrm{s} \max }=20 \mathrm{~m} / \mathrm{s}$
Since, as we know for the spring mass system
$\mathrm{v}_{\mathrm{s} \max }=\mathrm{A} \omega=\mathrm{A} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$
From equation (iii) and (iv), we get
$\mathrm{k}=\frac{\mathrm{v}_{\mathrm{s} \max }^{2} \times \mathrm{m}}{\mathrm{A}^{2}}$
$\mathrm{k}=\frac{(20)^{2} \times 0.10}{\left(50 \times 10^{-2}\right)^{2}}$
$\mathrm{k}=160 \mathrm{Nm}^{-1}$

AP EAMCET (21.04.2019) Shift-I

WAVES

172974 A siren placed at a railway platform is emitting a sound of frequency $5 \mathrm{kHz}$. A passenger sitting in a moving train $A$ records the frequency of the siren as $5.5 \mathrm{kHz}$. During his return journey by train $B$ he records the frequency of the siren as $6 \mathrm{kHz}$. The ratio of the speed of train $B$ to that of train $A$ is

1 $\frac{242}{252}$

2 2

3 $\frac{5}{6}$

4 $\frac{11}{6}$

AP EAMCET (20.04.2019) Shift-II

WAVES

172975 An observer moves towards a stationary source of sound with a speed $\frac{1^{\text {th }}}{5}$ of the speed of sound.
The wavelength and frequency of the waves emitted by the source are $\lambda$ and $f$ respectively. The apparent frequency and wavelength heard by the observer are respectively.

1 $1.2 f, \lambda$

2 $f, 1.2 \lambda$

3 $0.8 f, 0.8 \lambda$

4 $1.2 f, 1.2 \lambda$

AP EAMCET (20.04.2019) Shift-1

WAVES

172978 A car, sounding a horn of frequency $1000 \mathrm{~Hz}$, is moving directly towards a huge wall at a speed of $15 \mathrm{~m} / \mathrm{s}$. If the speed of sound is $340 \mathrm{~m} / \mathrm{s}$, then the frequency of echo heard by the driver is

1 $1046 \mathrm{~Hz}$

2 $954 \mathrm{~Hz}$

3 $1092 \mathrm{~Hz}$

4 $908 \mathrm{~Hz}$

CG PET -2018

WAVES

172973 A source of sound $S$ in the form of a block kept on a smooth horizontal surface is connected to a spring, as shown in the figure. If the spring oscillates with an amplitude of $50 \mathrm{~cm}$ along horizontal between the wall and the observer $o$, the maximum frequency heard by the observer is $12.5 \%$ more, than the minimum frequency heard by him. If the mass of the source of sound is $100 \mathrm{~g}$, the force constant of the spring is
(Speed of sound in air is $340 \mathrm{~ms}^{-1}$ )

1 $40 \mathrm{Nm}^{-1}$

2 $80 \mathrm{Nm}^{-1}$

3 $160 \mathrm{Nm}^{-1}$

4 $320 \mathrm{Nm}^{-1}$

Explanation:

C Given data,
Amplitude of spring $(A)=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$
Maximum frequency heard by the observer is $12.5 \%$ more than the minimum frequency
$\qquad \mathrm{n}_{\max }=1.125 \mathrm{n}_{\text {min }}$
$\text { Mass of source }(\mathrm{m})=100 \mathrm{~g}=\frac{100}{1000}=0.10 \mathrm{~kg}$
Speed of sound $\mathrm{v}=340 \mathrm{~m} / \mathrm{s}$

Apparent frequency $(\mathrm{n})=\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}$
For $\quad \mathrm{n}_{\max }$
$\text { For } \quad \mathrm{v}_{\mathrm{s}}=\mathrm{v}_{\mathrm{s} \max }$
$\mathrm{n}_{\min },$
$\mathrm{v}_{\mathrm{s}}=-\mathrm{v}_{\mathrm{s} \max }$
$\mathrm{n}_{\max }=\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}$
$\mathrm{n}_{\text {min }}=\frac{\eta_{\mathrm{o}} \mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }}$
On dividing Equation (i) \& (ii), we get-
$\frac{\mathrm{n}_{\text {max }}}{\mathrm{n}_{\text {min }}}=\frac{\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}}{\frac{\mathrm{n}_{\mathrm{o}} \mathrm{V}}{\mathrm{v}+\mathrm{v}_{\mathrm{s} \text { max }}}}$
$\frac{1.125 \mathrm{n}_{\text {min }}}{\mathrm{n}_{\text {min }}}=\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}$
$1.125\left(\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }\right)=\left(\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }\right)$
$0.125 \mathrm{v}=2.125 \mathrm{v}_{\mathrm{s} \max }$
$\mathrm{v}_{\mathrm{s} \max }=\frac{0.125 \times 340}{2.125}$
$\mathrm{v}_{\mathrm{s} \max }=20 \mathrm{~m} / \mathrm{s}$
Since, as we know for the spring mass system
$\mathrm{v}_{\mathrm{s} \max }=\mathrm{A} \omega=\mathrm{A} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$
From equation (iii) and (iv), we get
$\mathrm{k}=\frac{\mathrm{v}_{\mathrm{s} \max }^{2} \times \mathrm{m}}{\mathrm{A}^{2}}$
$\mathrm{k}=\frac{(20)^{2} \times 0.10}{\left(50 \times 10^{-2}\right)^{2}}$
$\mathrm{k}=160 \mathrm{Nm}^{-1}$

AP EAMCET (21.04.2019) Shift-I

WAVES

172974 A siren placed at a railway platform is emitting a sound of frequency $5 \mathrm{kHz}$. A passenger sitting in a moving train $A$ records the frequency of the siren as $5.5 \mathrm{kHz}$. During his return journey by train $B$ he records the frequency of the siren as $6 \mathrm{kHz}$. The ratio of the speed of train $B$ to that of train $A$ is

1 $\frac{242}{252}$

2 2

3 $\frac{5}{6}$

4 $\frac{11}{6}$

AP EAMCET (20.04.2019) Shift-II

WAVES

172975 An observer moves towards a stationary source of sound with a speed $\frac{1^{\text {th }}}{5}$ of the speed of sound.
The wavelength and frequency of the waves emitted by the source are $\lambda$ and $f$ respectively. The apparent frequency and wavelength heard by the observer are respectively.

1 $1.2 f, \lambda$

2 $f, 1.2 \lambda$

3 $0.8 f, 0.8 \lambda$

4 $1.2 f, 1.2 \lambda$

AP EAMCET (20.04.2019) Shift-1

WAVES

172978 A car, sounding a horn of frequency $1000 \mathrm{~Hz}$, is moving directly towards a huge wall at a speed of $15 \mathrm{~m} / \mathrm{s}$. If the speed of sound is $340 \mathrm{~m} / \mathrm{s}$, then the frequency of echo heard by the driver is

1 $1046 \mathrm{~Hz}$

2 $954 \mathrm{~Hz}$

3 $1092 \mathrm{~Hz}$

4 $908 \mathrm{~Hz}$

CG PET -2018

WAVES

172973 A source of sound $S$ in the form of a block kept on a smooth horizontal surface is connected to a spring, as shown in the figure. If the spring oscillates with an amplitude of $50 \mathrm{~cm}$ along horizontal between the wall and the observer $o$, the maximum frequency heard by the observer is $12.5 \%$ more, than the minimum frequency heard by him. If the mass of the source of sound is $100 \mathrm{~g}$, the force constant of the spring is
(Speed of sound in air is $340 \mathrm{~ms}^{-1}$ )

1 $40 \mathrm{Nm}^{-1}$

2 $80 \mathrm{Nm}^{-1}$

3 $160 \mathrm{Nm}^{-1}$

4 $320 \mathrm{Nm}^{-1}$

Explanation:

C Given data,
Amplitude of spring $(A)=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$
Maximum frequency heard by the observer is $12.5 \%$ more than the minimum frequency
$\qquad \mathrm{n}_{\max }=1.125 \mathrm{n}_{\text {min }}$
$\text { Mass of source }(\mathrm{m})=100 \mathrm{~g}=\frac{100}{1000}=0.10 \mathrm{~kg}$
Speed of sound $\mathrm{v}=340 \mathrm{~m} / \mathrm{s}$

Apparent frequency $(\mathrm{n})=\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}$
For $\quad \mathrm{n}_{\max }$
$\text { For } \quad \mathrm{v}_{\mathrm{s}}=\mathrm{v}_{\mathrm{s} \max }$
$\mathrm{n}_{\min },$
$\mathrm{v}_{\mathrm{s}}=-\mathrm{v}_{\mathrm{s} \max }$
$\mathrm{n}_{\max }=\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}$
$\mathrm{n}_{\text {min }}=\frac{\eta_{\mathrm{o}} \mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }}$
On dividing Equation (i) \& (ii), we get-
$\frac{\mathrm{n}_{\text {max }}}{\mathrm{n}_{\text {min }}}=\frac{\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}}{\frac{\mathrm{n}_{\mathrm{o}} \mathrm{V}}{\mathrm{v}+\mathrm{v}_{\mathrm{s} \text { max }}}}$
$\frac{1.125 \mathrm{n}_{\text {min }}}{\mathrm{n}_{\text {min }}}=\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}$
$1.125\left(\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }\right)=\left(\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }\right)$
$0.125 \mathrm{v}=2.125 \mathrm{v}_{\mathrm{s} \max }$
$\mathrm{v}_{\mathrm{s} \max }=\frac{0.125 \times 340}{2.125}$
$\mathrm{v}_{\mathrm{s} \max }=20 \mathrm{~m} / \mathrm{s}$
Since, as we know for the spring mass system
$\mathrm{v}_{\mathrm{s} \max }=\mathrm{A} \omega=\mathrm{A} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$
From equation (iii) and (iv), we get
$\mathrm{k}=\frac{\mathrm{v}_{\mathrm{s} \max }^{2} \times \mathrm{m}}{\mathrm{A}^{2}}$
$\mathrm{k}=\frac{(20)^{2} \times 0.10}{\left(50 \times 10^{-2}\right)^{2}}$
$\mathrm{k}=160 \mathrm{Nm}^{-1}$

AP EAMCET (21.04.2019) Shift-I

WAVES

172974 A siren placed at a railway platform is emitting a sound of frequency $5 \mathrm{kHz}$. A passenger sitting in a moving train $A$ records the frequency of the siren as $5.5 \mathrm{kHz}$. During his return journey by train $B$ he records the frequency of the siren as $6 \mathrm{kHz}$. The ratio of the speed of train $B$ to that of train $A$ is

1 $\frac{242}{252}$

2 2

3 $\frac{5}{6}$

4 $\frac{11}{6}$

AP EAMCET (20.04.2019) Shift-II

WAVES

172975 An observer moves towards a stationary source of sound with a speed $\frac{1^{\text {th }}}{5}$ of the speed of sound.
The wavelength and frequency of the waves emitted by the source are $\lambda$ and $f$ respectively. The apparent frequency and wavelength heard by the observer are respectively.

1 $1.2 f, \lambda$

2 $f, 1.2 \lambda$

3 $0.8 f, 0.8 \lambda$

4 $1.2 f, 1.2 \lambda$

AP EAMCET (20.04.2019) Shift-1

WAVES

172978 A car, sounding a horn of frequency $1000 \mathrm{~Hz}$, is moving directly towards a huge wall at a speed of $15 \mathrm{~m} / \mathrm{s}$. If the speed of sound is $340 \mathrm{~m} / \mathrm{s}$, then the frequency of echo heard by the driver is

1 $1046 \mathrm{~Hz}$

2 $954 \mathrm{~Hz}$

3 $1092 \mathrm{~Hz}$

4 $908 \mathrm{~Hz}$

CG PET -2018

WAVES

172973 A source of sound $S$ in the form of a block kept on a smooth horizontal surface is connected to a spring, as shown in the figure. If the spring oscillates with an amplitude of $50 \mathrm{~cm}$ along horizontal between the wall and the observer $o$, the maximum frequency heard by the observer is $12.5 \%$ more, than the minimum frequency heard by him. If the mass of the source of sound is $100 \mathrm{~g}$, the force constant of the spring is
(Speed of sound in air is $340 \mathrm{~ms}^{-1}$ )

1 $40 \mathrm{Nm}^{-1}$

2 $80 \mathrm{Nm}^{-1}$

3 $160 \mathrm{Nm}^{-1}$

4 $320 \mathrm{Nm}^{-1}$

Explanation:

C Given data,
Amplitude of spring $(A)=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$
Maximum frequency heard by the observer is $12.5 \%$ more than the minimum frequency
$\qquad \mathrm{n}_{\max }=1.125 \mathrm{n}_{\text {min }}$
$\text { Mass of source }(\mathrm{m})=100 \mathrm{~g}=\frac{100}{1000}=0.10 \mathrm{~kg}$
Speed of sound $\mathrm{v}=340 \mathrm{~m} / \mathrm{s}$

Apparent frequency $(\mathrm{n})=\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}$
For $\quad \mathrm{n}_{\max }$
$\text { For } \quad \mathrm{v}_{\mathrm{s}}=\mathrm{v}_{\mathrm{s} \max }$
$\mathrm{n}_{\min },$
$\mathrm{v}_{\mathrm{s}}=-\mathrm{v}_{\mathrm{s} \max }$
$\mathrm{n}_{\max }=\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}$
$\mathrm{n}_{\text {min }}=\frac{\eta_{\mathrm{o}} \mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }}$
On dividing Equation (i) \& (ii), we get-
$\frac{\mathrm{n}_{\text {max }}}{\mathrm{n}_{\text {min }}}=\frac{\frac{\mathrm{n}_{\mathrm{o}} \mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}}{\frac{\mathrm{n}_{\mathrm{o}} \mathrm{V}}{\mathrm{v}+\mathrm{v}_{\mathrm{s} \text { max }}}}$
$\frac{1.125 \mathrm{n}_{\text {min }}}{\mathrm{n}_{\text {min }}}=\frac{\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }}{\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }}$
$1.125\left(\mathrm{v}-\mathrm{v}_{\mathrm{s} \max }\right)=\left(\mathrm{v}+\mathrm{v}_{\mathrm{s} \max }\right)$
$0.125 \mathrm{v}=2.125 \mathrm{v}_{\mathrm{s} \max }$
$\mathrm{v}_{\mathrm{s} \max }=\frac{0.125 \times 340}{2.125}$
$\mathrm{v}_{\mathrm{s} \max }=20 \mathrm{~m} / \mathrm{s}$
Since, as we know for the spring mass system
$\mathrm{v}_{\mathrm{s} \max }=\mathrm{A} \omega=\mathrm{A} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$
From equation (iii) and (iv), we get
$\mathrm{k}=\frac{\mathrm{v}_{\mathrm{s} \max }^{2} \times \mathrm{m}}{\mathrm{A}^{2}}$
$\mathrm{k}=\frac{(20)^{2} \times 0.10}{\left(50 \times 10^{-2}\right)^{2}}$
$\mathrm{k}=160 \mathrm{Nm}^{-1}$

AP EAMCET (21.04.2019) Shift-I

WAVES

172974 A siren placed at a railway platform is emitting a sound of frequency $5 \mathrm{kHz}$. A passenger sitting in a moving train $A$ records the frequency of the siren as $5.5 \mathrm{kHz}$. During his return journey by train $B$ he records the frequency of the siren as $6 \mathrm{kHz}$. The ratio of the speed of train $B$ to that of train $A$ is

1 $\frac{242}{252}$

2 2

3 $\frac{5}{6}$

4 $\frac{11}{6}$

AP EAMCET (20.04.2019) Shift-II

WAVES

172975 An observer moves towards a stationary source of sound with a speed $\frac{1^{\text {th }}}{5}$ of the speed of sound.
The wavelength and frequency of the waves emitted by the source are $\lambda$ and $f$ respectively. The apparent frequency and wavelength heard by the observer are respectively.

1 $1.2 f, \lambda$

2 $f, 1.2 \lambda$

3 $0.8 f, 0.8 \lambda$

4 $1.2 f, 1.2 \lambda$

AP EAMCET (20.04.2019) Shift-1

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