173040
A car is moving towards a high cliff. The car drive sounds a horn of frequency $f$. The reflected sound heard by the driver has the frequency $2 f$. if $v$ be the velocity of sound, then the velocity of the car, in the same velocity units will be -
1 $\frac{\mathrm{v}}{\sqrt{2}}$
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{3}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
C We know that, Doppler's Effect $f^{\prime}=\left(\frac{v}{v-v_{s}}\right) f$ In case $I^{\text {st }}$ when the car is the source and at the cliff, our observer $\mathrm{f}^{\prime}$, $\therefore \quad \mathrm{f}^{\prime}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ In case II $^{\text {nd }}$ when cliff is now source, $f^{\prime \prime}=\left(\frac{v+v_{\circ}}{v}\right) f^{\prime}$ From equation (i), $\mathrm{f}^{\prime \prime}=\left(\frac{\mathrm{v}+\mathrm{v}_{\circ}}{\mathrm{v}}\right) \times\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $2 \mathrm{f}=\left(\frac{\mathrm{v}+\mathrm{v}_{\circ}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f} \quad\{\because \text { Given, } \mathrm{f} "=2 \mathrm{f}\}$ $2 \mathrm{v}-2 \mathrm{v}_{\circ}=\mathrm{v}+\mathrm{v}_{\circ} \quad\left(\because \mathrm{v}_{\mathrm{s}}=\mathrm{v}_{\circ}\right)$ $\mathrm{v}_{\circ}=\frac{\mathrm{v}}{3}$
BCECE-2013
WAVES
173041
A stationary source emits a whistle at a frequency of $200 \mathrm{~Hz}$. If the velocity of propagation of sound is $340 \mathrm{~m} / \mathrm{s}$ then the observed frequency, if the observer is moving away from the source at $25 \mathrm{~m} / \mathrm{s}$, will be :
1 $185 \mathrm{~Hz}$
2 $215 \mathrm{~Hz}$
3 $175 \mathrm{~Hz}$
4 $225 \mathrm{~Hz}$
Explanation:
A Given that, $\mathrm{v}_{\mathrm{o}}=25 \mathrm{~m} / \mathrm{s} \quad \mathrm{v}_{\mathrm{s}}=0, \mathrm{f}=200 \mathrm{~Hz}$, $\mathrm{v}=340 \mathrm{~m} / \mathrm{s}$ We know that, Observed frequency $\left(\mathrm{f}^{\prime}\right)=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $\therefore \quad \mathrm{f}^{\prime}=\left(\frac{340-25}{340-0}\right) 200$ $\mathrm{f}^{\prime}=\frac{315}{340} \times 200$ $\mathrm{f}^{\prime}=185.29 \square 185 \mathrm{~Hz}$
JIPMER-2014
WAVES
173042
Due to Doppler effect, the shift in wavelength observed is $0.1 \AA$, for a star producing wavelength $6000 \AA$. The velocity of recession of star will be-
1 $2.5 \mathrm{kms}^{-1}$
2 $10 \mathrm{kms}^{-1}$
3 $5 \mathrm{kms}^{-1}$
4 $20 \mathrm{kms}^{-1}$
Explanation:
C Given that, $\Delta \lambda=0.1 \AA$ $\lambda=6000 \AA$ Light speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know the formula for Doppler shift, $\frac{\Delta \lambda}{\lambda}=\frac{\mathrm{v}}{\mathrm{c}}$ $\mathrm{v}=\frac{\Delta \lambda}{\lambda} \times \mathrm{c}=\frac{0.1}{6000} \times 3 \times 10^{8}$ $\mathrm{v}=0.5 \times 10^{4}$ $\mathrm{v}=5 \mathrm{~km} / \mathrm{s}$
BCECE-2009
WAVES
173043
When both the listener and source are moving towards each other, then which of the following is true regarding frequency and wavelength of wave observed by the observer?
1 More frequency, less wavelength
2 More frequency, more wavelength
3 Less frequency, less wavelength
4 More frequency, constant wavelength
Explanation:
A According to Doppler effect $f^{\prime}=\left(\frac{v+v_{\circ}}{v-v_{s}}\right) f_{\circ}$ Velocity of observer $\left(\mathrm{v}_{\mathrm{o}}\right)=$ positive, when the observer is moving towards the source. Velocity of source $\left(\mathrm{v}_{\mathrm{s}}\right)=$ Positive, when the source is moving toward the observer $\mathrm{f}^{\prime}>\mathrm{f}_{\circ}$ $\lambda>\lambda_{\circ}(\text { since } \mathrm{f}=1 / \lambda)$ So, listener, listens more frequency and observe less wavelength.
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WAVES
173040
A car is moving towards a high cliff. The car drive sounds a horn of frequency $f$. The reflected sound heard by the driver has the frequency $2 f$. if $v$ be the velocity of sound, then the velocity of the car, in the same velocity units will be -
1 $\frac{\mathrm{v}}{\sqrt{2}}$
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{3}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
C We know that, Doppler's Effect $f^{\prime}=\left(\frac{v}{v-v_{s}}\right) f$ In case $I^{\text {st }}$ when the car is the source and at the cliff, our observer $\mathrm{f}^{\prime}$, $\therefore \quad \mathrm{f}^{\prime}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ In case II $^{\text {nd }}$ when cliff is now source, $f^{\prime \prime}=\left(\frac{v+v_{\circ}}{v}\right) f^{\prime}$ From equation (i), $\mathrm{f}^{\prime \prime}=\left(\frac{\mathrm{v}+\mathrm{v}_{\circ}}{\mathrm{v}}\right) \times\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $2 \mathrm{f}=\left(\frac{\mathrm{v}+\mathrm{v}_{\circ}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f} \quad\{\because \text { Given, } \mathrm{f} "=2 \mathrm{f}\}$ $2 \mathrm{v}-2 \mathrm{v}_{\circ}=\mathrm{v}+\mathrm{v}_{\circ} \quad\left(\because \mathrm{v}_{\mathrm{s}}=\mathrm{v}_{\circ}\right)$ $\mathrm{v}_{\circ}=\frac{\mathrm{v}}{3}$
BCECE-2013
WAVES
173041
A stationary source emits a whistle at a frequency of $200 \mathrm{~Hz}$. If the velocity of propagation of sound is $340 \mathrm{~m} / \mathrm{s}$ then the observed frequency, if the observer is moving away from the source at $25 \mathrm{~m} / \mathrm{s}$, will be :
1 $185 \mathrm{~Hz}$
2 $215 \mathrm{~Hz}$
3 $175 \mathrm{~Hz}$
4 $225 \mathrm{~Hz}$
Explanation:
A Given that, $\mathrm{v}_{\mathrm{o}}=25 \mathrm{~m} / \mathrm{s} \quad \mathrm{v}_{\mathrm{s}}=0, \mathrm{f}=200 \mathrm{~Hz}$, $\mathrm{v}=340 \mathrm{~m} / \mathrm{s}$ We know that, Observed frequency $\left(\mathrm{f}^{\prime}\right)=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $\therefore \quad \mathrm{f}^{\prime}=\left(\frac{340-25}{340-0}\right) 200$ $\mathrm{f}^{\prime}=\frac{315}{340} \times 200$ $\mathrm{f}^{\prime}=185.29 \square 185 \mathrm{~Hz}$
JIPMER-2014
WAVES
173042
Due to Doppler effect, the shift in wavelength observed is $0.1 \AA$, for a star producing wavelength $6000 \AA$. The velocity of recession of star will be-
1 $2.5 \mathrm{kms}^{-1}$
2 $10 \mathrm{kms}^{-1}$
3 $5 \mathrm{kms}^{-1}$
4 $20 \mathrm{kms}^{-1}$
Explanation:
C Given that, $\Delta \lambda=0.1 \AA$ $\lambda=6000 \AA$ Light speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know the formula for Doppler shift, $\frac{\Delta \lambda}{\lambda}=\frac{\mathrm{v}}{\mathrm{c}}$ $\mathrm{v}=\frac{\Delta \lambda}{\lambda} \times \mathrm{c}=\frac{0.1}{6000} \times 3 \times 10^{8}$ $\mathrm{v}=0.5 \times 10^{4}$ $\mathrm{v}=5 \mathrm{~km} / \mathrm{s}$
BCECE-2009
WAVES
173043
When both the listener and source are moving towards each other, then which of the following is true regarding frequency and wavelength of wave observed by the observer?
1 More frequency, less wavelength
2 More frequency, more wavelength
3 Less frequency, less wavelength
4 More frequency, constant wavelength
Explanation:
A According to Doppler effect $f^{\prime}=\left(\frac{v+v_{\circ}}{v-v_{s}}\right) f_{\circ}$ Velocity of observer $\left(\mathrm{v}_{\mathrm{o}}\right)=$ positive, when the observer is moving towards the source. Velocity of source $\left(\mathrm{v}_{\mathrm{s}}\right)=$ Positive, when the source is moving toward the observer $\mathrm{f}^{\prime}>\mathrm{f}_{\circ}$ $\lambda>\lambda_{\circ}(\text { since } \mathrm{f}=1 / \lambda)$ So, listener, listens more frequency and observe less wavelength.
173040
A car is moving towards a high cliff. The car drive sounds a horn of frequency $f$. The reflected sound heard by the driver has the frequency $2 f$. if $v$ be the velocity of sound, then the velocity of the car, in the same velocity units will be -
1 $\frac{\mathrm{v}}{\sqrt{2}}$
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{3}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
C We know that, Doppler's Effect $f^{\prime}=\left(\frac{v}{v-v_{s}}\right) f$ In case $I^{\text {st }}$ when the car is the source and at the cliff, our observer $\mathrm{f}^{\prime}$, $\therefore \quad \mathrm{f}^{\prime}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ In case II $^{\text {nd }}$ when cliff is now source, $f^{\prime \prime}=\left(\frac{v+v_{\circ}}{v}\right) f^{\prime}$ From equation (i), $\mathrm{f}^{\prime \prime}=\left(\frac{\mathrm{v}+\mathrm{v}_{\circ}}{\mathrm{v}}\right) \times\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $2 \mathrm{f}=\left(\frac{\mathrm{v}+\mathrm{v}_{\circ}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f} \quad\{\because \text { Given, } \mathrm{f} "=2 \mathrm{f}\}$ $2 \mathrm{v}-2 \mathrm{v}_{\circ}=\mathrm{v}+\mathrm{v}_{\circ} \quad\left(\because \mathrm{v}_{\mathrm{s}}=\mathrm{v}_{\circ}\right)$ $\mathrm{v}_{\circ}=\frac{\mathrm{v}}{3}$
BCECE-2013
WAVES
173041
A stationary source emits a whistle at a frequency of $200 \mathrm{~Hz}$. If the velocity of propagation of sound is $340 \mathrm{~m} / \mathrm{s}$ then the observed frequency, if the observer is moving away from the source at $25 \mathrm{~m} / \mathrm{s}$, will be :
1 $185 \mathrm{~Hz}$
2 $215 \mathrm{~Hz}$
3 $175 \mathrm{~Hz}$
4 $225 \mathrm{~Hz}$
Explanation:
A Given that, $\mathrm{v}_{\mathrm{o}}=25 \mathrm{~m} / \mathrm{s} \quad \mathrm{v}_{\mathrm{s}}=0, \mathrm{f}=200 \mathrm{~Hz}$, $\mathrm{v}=340 \mathrm{~m} / \mathrm{s}$ We know that, Observed frequency $\left(\mathrm{f}^{\prime}\right)=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $\therefore \quad \mathrm{f}^{\prime}=\left(\frac{340-25}{340-0}\right) 200$ $\mathrm{f}^{\prime}=\frac{315}{340} \times 200$ $\mathrm{f}^{\prime}=185.29 \square 185 \mathrm{~Hz}$
JIPMER-2014
WAVES
173042
Due to Doppler effect, the shift in wavelength observed is $0.1 \AA$, for a star producing wavelength $6000 \AA$. The velocity of recession of star will be-
1 $2.5 \mathrm{kms}^{-1}$
2 $10 \mathrm{kms}^{-1}$
3 $5 \mathrm{kms}^{-1}$
4 $20 \mathrm{kms}^{-1}$
Explanation:
C Given that, $\Delta \lambda=0.1 \AA$ $\lambda=6000 \AA$ Light speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know the formula for Doppler shift, $\frac{\Delta \lambda}{\lambda}=\frac{\mathrm{v}}{\mathrm{c}}$ $\mathrm{v}=\frac{\Delta \lambda}{\lambda} \times \mathrm{c}=\frac{0.1}{6000} \times 3 \times 10^{8}$ $\mathrm{v}=0.5 \times 10^{4}$ $\mathrm{v}=5 \mathrm{~km} / \mathrm{s}$
BCECE-2009
WAVES
173043
When both the listener and source are moving towards each other, then which of the following is true regarding frequency and wavelength of wave observed by the observer?
1 More frequency, less wavelength
2 More frequency, more wavelength
3 Less frequency, less wavelength
4 More frequency, constant wavelength
Explanation:
A According to Doppler effect $f^{\prime}=\left(\frac{v+v_{\circ}}{v-v_{s}}\right) f_{\circ}$ Velocity of observer $\left(\mathrm{v}_{\mathrm{o}}\right)=$ positive, when the observer is moving towards the source. Velocity of source $\left(\mathrm{v}_{\mathrm{s}}\right)=$ Positive, when the source is moving toward the observer $\mathrm{f}^{\prime}>\mathrm{f}_{\circ}$ $\lambda>\lambda_{\circ}(\text { since } \mathrm{f}=1 / \lambda)$ So, listener, listens more frequency and observe less wavelength.
173040
A car is moving towards a high cliff. The car drive sounds a horn of frequency $f$. The reflected sound heard by the driver has the frequency $2 f$. if $v$ be the velocity of sound, then the velocity of the car, in the same velocity units will be -
1 $\frac{\mathrm{v}}{\sqrt{2}}$
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{3}$
4 $\frac{\mathrm{V}}{4}$
Explanation:
C We know that, Doppler's Effect $f^{\prime}=\left(\frac{v}{v-v_{s}}\right) f$ In case $I^{\text {st }}$ when the car is the source and at the cliff, our observer $\mathrm{f}^{\prime}$, $\therefore \quad \mathrm{f}^{\prime}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ In case II $^{\text {nd }}$ when cliff is now source, $f^{\prime \prime}=\left(\frac{v+v_{\circ}}{v}\right) f^{\prime}$ From equation (i), $\mathrm{f}^{\prime \prime}=\left(\frac{\mathrm{v}+\mathrm{v}_{\circ}}{\mathrm{v}}\right) \times\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $2 \mathrm{f}=\left(\frac{\mathrm{v}+\mathrm{v}_{\circ}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f} \quad\{\because \text { Given, } \mathrm{f} "=2 \mathrm{f}\}$ $2 \mathrm{v}-2 \mathrm{v}_{\circ}=\mathrm{v}+\mathrm{v}_{\circ} \quad\left(\because \mathrm{v}_{\mathrm{s}}=\mathrm{v}_{\circ}\right)$ $\mathrm{v}_{\circ}=\frac{\mathrm{v}}{3}$
BCECE-2013
WAVES
173041
A stationary source emits a whistle at a frequency of $200 \mathrm{~Hz}$. If the velocity of propagation of sound is $340 \mathrm{~m} / \mathrm{s}$ then the observed frequency, if the observer is moving away from the source at $25 \mathrm{~m} / \mathrm{s}$, will be :
1 $185 \mathrm{~Hz}$
2 $215 \mathrm{~Hz}$
3 $175 \mathrm{~Hz}$
4 $225 \mathrm{~Hz}$
Explanation:
A Given that, $\mathrm{v}_{\mathrm{o}}=25 \mathrm{~m} / \mathrm{s} \quad \mathrm{v}_{\mathrm{s}}=0, \mathrm{f}=200 \mathrm{~Hz}$, $\mathrm{v}=340 \mathrm{~m} / \mathrm{s}$ We know that, Observed frequency $\left(\mathrm{f}^{\prime}\right)=\left(\frac{\mathrm{v}-\mathrm{v}_{\mathrm{o}}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{f}$ $\therefore \quad \mathrm{f}^{\prime}=\left(\frac{340-25}{340-0}\right) 200$ $\mathrm{f}^{\prime}=\frac{315}{340} \times 200$ $\mathrm{f}^{\prime}=185.29 \square 185 \mathrm{~Hz}$
JIPMER-2014
WAVES
173042
Due to Doppler effect, the shift in wavelength observed is $0.1 \AA$, for a star producing wavelength $6000 \AA$. The velocity of recession of star will be-
1 $2.5 \mathrm{kms}^{-1}$
2 $10 \mathrm{kms}^{-1}$
3 $5 \mathrm{kms}^{-1}$
4 $20 \mathrm{kms}^{-1}$
Explanation:
C Given that, $\Delta \lambda=0.1 \AA$ $\lambda=6000 \AA$ Light speed $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ We know the formula for Doppler shift, $\frac{\Delta \lambda}{\lambda}=\frac{\mathrm{v}}{\mathrm{c}}$ $\mathrm{v}=\frac{\Delta \lambda}{\lambda} \times \mathrm{c}=\frac{0.1}{6000} \times 3 \times 10^{8}$ $\mathrm{v}=0.5 \times 10^{4}$ $\mathrm{v}=5 \mathrm{~km} / \mathrm{s}$
BCECE-2009
WAVES
173043
When both the listener and source are moving towards each other, then which of the following is true regarding frequency and wavelength of wave observed by the observer?
1 More frequency, less wavelength
2 More frequency, more wavelength
3 Less frequency, less wavelength
4 More frequency, constant wavelength
Explanation:
A According to Doppler effect $f^{\prime}=\left(\frac{v+v_{\circ}}{v-v_{s}}\right) f_{\circ}$ Velocity of observer $\left(\mathrm{v}_{\mathrm{o}}\right)=$ positive, when the observer is moving towards the source. Velocity of source $\left(\mathrm{v}_{\mathrm{s}}\right)=$ Positive, when the source is moving toward the observer $\mathrm{f}^{\prime}>\mathrm{f}_{\circ}$ $\lambda>\lambda_{\circ}(\text { since } \mathrm{f}=1 / \lambda)$ So, listener, listens more frequency and observe less wavelength.