A Generalised formula for apparent frequency is $\mathrm{f}^{\prime}=\left(\frac{\mathrm{v} \pm \mathrm{v}_{\mathrm{o}}}{\mathrm{v} \pm \mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}_{\mathrm{o}}$ According to the question source is stationary, $\mathrm{v}_{\mathrm{s}}=0$ $\mathrm{f}_{1}{ }_{1}=\left(\frac{\mathrm{v} \pm \mathrm{v}_{\circ}}{\mathrm{v}}\right) \mathrm{f}_{\circ}$ Observer is moving towards the source $\Rightarrow \quad \mathrm{f}^{\prime}{ }_{2}=\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}}\right) \mathrm{f}$ Where, $\mathrm{v}=$ speed of the wave $v_{\circ}=\text { speed of the observer }$ $\mathrm{f}=\text { original frequency of the source }$
Manipal UGET -2020
WAVES
172951
A whistle of frequency $1000 \mathrm{~Hz}$ is sounded on a car travelling towards a cliff with velocity of 18 $\mathrm{ms}^{-1}$ normal to the cliff. If $\mathrm{c}=330 \mathrm{~ms}^{-1}$, then the apparent frequency of the echo as heard by the car driver is nearly
1 $1115 \mathrm{~Hz}$
2 $115 \mathrm{~Hz}$
3 $67 \mathrm{~Hz}$
4 $47.2 \mathrm{~Hz}$
Explanation:
A Given that, substitute $330 \mathrm{~m} / \mathrm{s}$ for $\mathrm{v}$, $\mathrm{f}=1000 \mathrm{~Hz}$, Velocity of an echo of sound $\left(\mathrm{v}_{\mathrm{s}}\right)=18 \mathrm{~m} / \mathrm{s}$, Velocity of an observer $\left(\mathrm{v}_{\mathrm{o}}\right)=18 \mathrm{~m} / \mathrm{s}$ We know that, Doppler formula $\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $\mathrm{f}^{\prime}=\text { Apparent frequency }$ $\mathrm{f}=\text { Original frequency }$ $\mathrm{v}=\text { Velocity of sound }$ $\mathrm{v}_{\circ}=\text { Velocity of an observer }$ $\mathrm{v}_{\mathrm{s}}=\text { Velocity of the source }$ $\mathrm{f}^{\prime}=1000\left(\frac{330+18}{330-18}\right)$ $\mathrm{f}^{\prime}=1000 \times \frac{348}{312}$ $\mathrm{f}^{\prime}=1115 \mathrm{~Hz}$
BITSAT-2020
WAVES
172953
A source is stationary and the observer is in motion along a line joining the source and the observer. If the frequency heard by the observer is $1 \%$ higher than the true frequency, the ratio of velocity of the observer and that of sound in air is
1 $1: 100$
2 $2: 100$
3 $3: 100$
4 $1: 10$
Explanation:
A Given that, $v_{\mathrm{s}}=0$ Let, velocity of observer $=v_{0}$ Sound in air $=\mathrm{v}$ True frequency of sound $=\mathrm{f}_{\mathrm{o}}$ Heard frequency $(f)=1 \%$ higher than true frequency. $\mathrm{f}=1.01 \mathrm{f}_{\mathrm{o}}$ According to the Doppler's effect, if source is stationary and observer is moving towards source along a line joining them. Then, $\text { Heard frequency, } f=\left(\frac{v+v_{o}}{v}\right) f_{o}$ $\frac{f}{f_{o}}=\frac{v+v_{o}}{v}$ $1.01=\frac{v+v_{o}}{v}$ $1.01 \mathrm{v}=v+v_{o}$ $0.01 \mathrm{v}=v_{o}$ $\frac{v_{o}}{v}=0.01=\frac{1}{100}$
AP EAMCET (23.09.2020) Shift-I
WAVES
172955
The frequency of a note emitted by a source changes by $10 \%$ as it moves away from a stationary observer. If it moves towards the stationary observer with the same speed. the apparent change in the frequency is
1 $10 \%$
2 $7.5 \%$
3 $12.5 \%$
4 $20 \%$
Explanation:
C Frequency $10 \%$ change when it move away from stationary observer $\left(\frac{90}{100}\right) \mathrm{f}=\mathrm{f}\left(\frac{\mathrm{v}}{\mathrm{v} \times \mathrm{v}_{\mathrm{s}}}\right)$ $\mathrm{v}_{\mathrm{s}}=\frac{1}{9} \mathrm{v}$ Therefore the percentage change when its approach towards the observer $\left(\frac{100-\mathrm{x}}{100}\right) \times \mathrm{f}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \times \mathrm{f}$ $\left(1-\frac{\mathrm{x}}{100}\right)=\frac{\mathrm{v}}{\mathrm{v}-\frac{1}{9} \mathrm{v}}=\frac{9}{8}$ $\frac{\mathrm{x}}{100}=1-\frac{9}{8}$ $\mathrm{x}=-\frac{1}{8} \times 100$ $\mathrm{x}=12.5 \%, \text { decreases. }$
A Generalised formula for apparent frequency is $\mathrm{f}^{\prime}=\left(\frac{\mathrm{v} \pm \mathrm{v}_{\mathrm{o}}}{\mathrm{v} \pm \mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}_{\mathrm{o}}$ According to the question source is stationary, $\mathrm{v}_{\mathrm{s}}=0$ $\mathrm{f}_{1}{ }_{1}=\left(\frac{\mathrm{v} \pm \mathrm{v}_{\circ}}{\mathrm{v}}\right) \mathrm{f}_{\circ}$ Observer is moving towards the source $\Rightarrow \quad \mathrm{f}^{\prime}{ }_{2}=\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}}\right) \mathrm{f}$ Where, $\mathrm{v}=$ speed of the wave $v_{\circ}=\text { speed of the observer }$ $\mathrm{f}=\text { original frequency of the source }$
Manipal UGET -2020
WAVES
172951
A whistle of frequency $1000 \mathrm{~Hz}$ is sounded on a car travelling towards a cliff with velocity of 18 $\mathrm{ms}^{-1}$ normal to the cliff. If $\mathrm{c}=330 \mathrm{~ms}^{-1}$, then the apparent frequency of the echo as heard by the car driver is nearly
1 $1115 \mathrm{~Hz}$
2 $115 \mathrm{~Hz}$
3 $67 \mathrm{~Hz}$
4 $47.2 \mathrm{~Hz}$
Explanation:
A Given that, substitute $330 \mathrm{~m} / \mathrm{s}$ for $\mathrm{v}$, $\mathrm{f}=1000 \mathrm{~Hz}$, Velocity of an echo of sound $\left(\mathrm{v}_{\mathrm{s}}\right)=18 \mathrm{~m} / \mathrm{s}$, Velocity of an observer $\left(\mathrm{v}_{\mathrm{o}}\right)=18 \mathrm{~m} / \mathrm{s}$ We know that, Doppler formula $\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $\mathrm{f}^{\prime}=\text { Apparent frequency }$ $\mathrm{f}=\text { Original frequency }$ $\mathrm{v}=\text { Velocity of sound }$ $\mathrm{v}_{\circ}=\text { Velocity of an observer }$ $\mathrm{v}_{\mathrm{s}}=\text { Velocity of the source }$ $\mathrm{f}^{\prime}=1000\left(\frac{330+18}{330-18}\right)$ $\mathrm{f}^{\prime}=1000 \times \frac{348}{312}$ $\mathrm{f}^{\prime}=1115 \mathrm{~Hz}$
BITSAT-2020
WAVES
172953
A source is stationary and the observer is in motion along a line joining the source and the observer. If the frequency heard by the observer is $1 \%$ higher than the true frequency, the ratio of velocity of the observer and that of sound in air is
1 $1: 100$
2 $2: 100$
3 $3: 100$
4 $1: 10$
Explanation:
A Given that, $v_{\mathrm{s}}=0$ Let, velocity of observer $=v_{0}$ Sound in air $=\mathrm{v}$ True frequency of sound $=\mathrm{f}_{\mathrm{o}}$ Heard frequency $(f)=1 \%$ higher than true frequency. $\mathrm{f}=1.01 \mathrm{f}_{\mathrm{o}}$ According to the Doppler's effect, if source is stationary and observer is moving towards source along a line joining them. Then, $\text { Heard frequency, } f=\left(\frac{v+v_{o}}{v}\right) f_{o}$ $\frac{f}{f_{o}}=\frac{v+v_{o}}{v}$ $1.01=\frac{v+v_{o}}{v}$ $1.01 \mathrm{v}=v+v_{o}$ $0.01 \mathrm{v}=v_{o}$ $\frac{v_{o}}{v}=0.01=\frac{1}{100}$
AP EAMCET (23.09.2020) Shift-I
WAVES
172955
The frequency of a note emitted by a source changes by $10 \%$ as it moves away from a stationary observer. If it moves towards the stationary observer with the same speed. the apparent change in the frequency is
1 $10 \%$
2 $7.5 \%$
3 $12.5 \%$
4 $20 \%$
Explanation:
C Frequency $10 \%$ change when it move away from stationary observer $\left(\frac{90}{100}\right) \mathrm{f}=\mathrm{f}\left(\frac{\mathrm{v}}{\mathrm{v} \times \mathrm{v}_{\mathrm{s}}}\right)$ $\mathrm{v}_{\mathrm{s}}=\frac{1}{9} \mathrm{v}$ Therefore the percentage change when its approach towards the observer $\left(\frac{100-\mathrm{x}}{100}\right) \times \mathrm{f}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \times \mathrm{f}$ $\left(1-\frac{\mathrm{x}}{100}\right)=\frac{\mathrm{v}}{\mathrm{v}-\frac{1}{9} \mathrm{v}}=\frac{9}{8}$ $\frac{\mathrm{x}}{100}=1-\frac{9}{8}$ $\mathrm{x}=-\frac{1}{8} \times 100$ $\mathrm{x}=12.5 \%, \text { decreases. }$
A Generalised formula for apparent frequency is $\mathrm{f}^{\prime}=\left(\frac{\mathrm{v} \pm \mathrm{v}_{\mathrm{o}}}{\mathrm{v} \pm \mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}_{\mathrm{o}}$ According to the question source is stationary, $\mathrm{v}_{\mathrm{s}}=0$ $\mathrm{f}_{1}{ }_{1}=\left(\frac{\mathrm{v} \pm \mathrm{v}_{\circ}}{\mathrm{v}}\right) \mathrm{f}_{\circ}$ Observer is moving towards the source $\Rightarrow \quad \mathrm{f}^{\prime}{ }_{2}=\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}}\right) \mathrm{f}$ Where, $\mathrm{v}=$ speed of the wave $v_{\circ}=\text { speed of the observer }$ $\mathrm{f}=\text { original frequency of the source }$
Manipal UGET -2020
WAVES
172951
A whistle of frequency $1000 \mathrm{~Hz}$ is sounded on a car travelling towards a cliff with velocity of 18 $\mathrm{ms}^{-1}$ normal to the cliff. If $\mathrm{c}=330 \mathrm{~ms}^{-1}$, then the apparent frequency of the echo as heard by the car driver is nearly
1 $1115 \mathrm{~Hz}$
2 $115 \mathrm{~Hz}$
3 $67 \mathrm{~Hz}$
4 $47.2 \mathrm{~Hz}$
Explanation:
A Given that, substitute $330 \mathrm{~m} / \mathrm{s}$ for $\mathrm{v}$, $\mathrm{f}=1000 \mathrm{~Hz}$, Velocity of an echo of sound $\left(\mathrm{v}_{\mathrm{s}}\right)=18 \mathrm{~m} / \mathrm{s}$, Velocity of an observer $\left(\mathrm{v}_{\mathrm{o}}\right)=18 \mathrm{~m} / \mathrm{s}$ We know that, Doppler formula $\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $\mathrm{f}^{\prime}=\text { Apparent frequency }$ $\mathrm{f}=\text { Original frequency }$ $\mathrm{v}=\text { Velocity of sound }$ $\mathrm{v}_{\circ}=\text { Velocity of an observer }$ $\mathrm{v}_{\mathrm{s}}=\text { Velocity of the source }$ $\mathrm{f}^{\prime}=1000\left(\frac{330+18}{330-18}\right)$ $\mathrm{f}^{\prime}=1000 \times \frac{348}{312}$ $\mathrm{f}^{\prime}=1115 \mathrm{~Hz}$
BITSAT-2020
WAVES
172953
A source is stationary and the observer is in motion along a line joining the source and the observer. If the frequency heard by the observer is $1 \%$ higher than the true frequency, the ratio of velocity of the observer and that of sound in air is
1 $1: 100$
2 $2: 100$
3 $3: 100$
4 $1: 10$
Explanation:
A Given that, $v_{\mathrm{s}}=0$ Let, velocity of observer $=v_{0}$ Sound in air $=\mathrm{v}$ True frequency of sound $=\mathrm{f}_{\mathrm{o}}$ Heard frequency $(f)=1 \%$ higher than true frequency. $\mathrm{f}=1.01 \mathrm{f}_{\mathrm{o}}$ According to the Doppler's effect, if source is stationary and observer is moving towards source along a line joining them. Then, $\text { Heard frequency, } f=\left(\frac{v+v_{o}}{v}\right) f_{o}$ $\frac{f}{f_{o}}=\frac{v+v_{o}}{v}$ $1.01=\frac{v+v_{o}}{v}$ $1.01 \mathrm{v}=v+v_{o}$ $0.01 \mathrm{v}=v_{o}$ $\frac{v_{o}}{v}=0.01=\frac{1}{100}$
AP EAMCET (23.09.2020) Shift-I
WAVES
172955
The frequency of a note emitted by a source changes by $10 \%$ as it moves away from a stationary observer. If it moves towards the stationary observer with the same speed. the apparent change in the frequency is
1 $10 \%$
2 $7.5 \%$
3 $12.5 \%$
4 $20 \%$
Explanation:
C Frequency $10 \%$ change when it move away from stationary observer $\left(\frac{90}{100}\right) \mathrm{f}=\mathrm{f}\left(\frac{\mathrm{v}}{\mathrm{v} \times \mathrm{v}_{\mathrm{s}}}\right)$ $\mathrm{v}_{\mathrm{s}}=\frac{1}{9} \mathrm{v}$ Therefore the percentage change when its approach towards the observer $\left(\frac{100-\mathrm{x}}{100}\right) \times \mathrm{f}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \times \mathrm{f}$ $\left(1-\frac{\mathrm{x}}{100}\right)=\frac{\mathrm{v}}{\mathrm{v}-\frac{1}{9} \mathrm{v}}=\frac{9}{8}$ $\frac{\mathrm{x}}{100}=1-\frac{9}{8}$ $\mathrm{x}=-\frac{1}{8} \times 100$ $\mathrm{x}=12.5 \%, \text { decreases. }$
A Generalised formula for apparent frequency is $\mathrm{f}^{\prime}=\left(\frac{\mathrm{v} \pm \mathrm{v}_{\mathrm{o}}}{\mathrm{v} \pm \mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}_{\mathrm{o}}$ According to the question source is stationary, $\mathrm{v}_{\mathrm{s}}=0$ $\mathrm{f}_{1}{ }_{1}=\left(\frac{\mathrm{v} \pm \mathrm{v}_{\circ}}{\mathrm{v}}\right) \mathrm{f}_{\circ}$ Observer is moving towards the source $\Rightarrow \quad \mathrm{f}^{\prime}{ }_{2}=\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}}\right) \mathrm{f}$ Where, $\mathrm{v}=$ speed of the wave $v_{\circ}=\text { speed of the observer }$ $\mathrm{f}=\text { original frequency of the source }$
Manipal UGET -2020
WAVES
172951
A whistle of frequency $1000 \mathrm{~Hz}$ is sounded on a car travelling towards a cliff with velocity of 18 $\mathrm{ms}^{-1}$ normal to the cliff. If $\mathrm{c}=330 \mathrm{~ms}^{-1}$, then the apparent frequency of the echo as heard by the car driver is nearly
1 $1115 \mathrm{~Hz}$
2 $115 \mathrm{~Hz}$
3 $67 \mathrm{~Hz}$
4 $47.2 \mathrm{~Hz}$
Explanation:
A Given that, substitute $330 \mathrm{~m} / \mathrm{s}$ for $\mathrm{v}$, $\mathrm{f}=1000 \mathrm{~Hz}$, Velocity of an echo of sound $\left(\mathrm{v}_{\mathrm{s}}\right)=18 \mathrm{~m} / \mathrm{s}$, Velocity of an observer $\left(\mathrm{v}_{\mathrm{o}}\right)=18 \mathrm{~m} / \mathrm{s}$ We know that, Doppler formula $\mathrm{f}^{\prime}=\mathrm{f}\left(\frac{\mathrm{v}+\mathrm{v}_{\mathrm{o}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right)$ $\mathrm{f}^{\prime}=\text { Apparent frequency }$ $\mathrm{f}=\text { Original frequency }$ $\mathrm{v}=\text { Velocity of sound }$ $\mathrm{v}_{\circ}=\text { Velocity of an observer }$ $\mathrm{v}_{\mathrm{s}}=\text { Velocity of the source }$ $\mathrm{f}^{\prime}=1000\left(\frac{330+18}{330-18}\right)$ $\mathrm{f}^{\prime}=1000 \times \frac{348}{312}$ $\mathrm{f}^{\prime}=1115 \mathrm{~Hz}$
BITSAT-2020
WAVES
172953
A source is stationary and the observer is in motion along a line joining the source and the observer. If the frequency heard by the observer is $1 \%$ higher than the true frequency, the ratio of velocity of the observer and that of sound in air is
1 $1: 100$
2 $2: 100$
3 $3: 100$
4 $1: 10$
Explanation:
A Given that, $v_{\mathrm{s}}=0$ Let, velocity of observer $=v_{0}$ Sound in air $=\mathrm{v}$ True frequency of sound $=\mathrm{f}_{\mathrm{o}}$ Heard frequency $(f)=1 \%$ higher than true frequency. $\mathrm{f}=1.01 \mathrm{f}_{\mathrm{o}}$ According to the Doppler's effect, if source is stationary and observer is moving towards source along a line joining them. Then, $\text { Heard frequency, } f=\left(\frac{v+v_{o}}{v}\right) f_{o}$ $\frac{f}{f_{o}}=\frac{v+v_{o}}{v}$ $1.01=\frac{v+v_{o}}{v}$ $1.01 \mathrm{v}=v+v_{o}$ $0.01 \mathrm{v}=v_{o}$ $\frac{v_{o}}{v}=0.01=\frac{1}{100}$
AP EAMCET (23.09.2020) Shift-I
WAVES
172955
The frequency of a note emitted by a source changes by $10 \%$ as it moves away from a stationary observer. If it moves towards the stationary observer with the same speed. the apparent change in the frequency is
1 $10 \%$
2 $7.5 \%$
3 $12.5 \%$
4 $20 \%$
Explanation:
C Frequency $10 \%$ change when it move away from stationary observer $\left(\frac{90}{100}\right) \mathrm{f}=\mathrm{f}\left(\frac{\mathrm{v}}{\mathrm{v} \times \mathrm{v}_{\mathrm{s}}}\right)$ $\mathrm{v}_{\mathrm{s}}=\frac{1}{9} \mathrm{v}$ Therefore the percentage change when its approach towards the observer $\left(\frac{100-\mathrm{x}}{100}\right) \times \mathrm{f}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \times \mathrm{f}$ $\left(1-\frac{\mathrm{x}}{100}\right)=\frac{\mathrm{v}}{\mathrm{v}-\frac{1}{9} \mathrm{v}}=\frac{9}{8}$ $\frac{\mathrm{x}}{100}=1-\frac{9}{8}$ $\mathrm{x}=-\frac{1}{8} \times 100$ $\mathrm{x}=12.5 \%, \text { decreases. }$