172364
Two harmonic travelling waves are described by the equations $y_{1}=a \sin (k x-\omega t)$ and $y_{2}=a$ $\sin (-k x+\omega t+\phi)$. The amplitude of the superimposed wave is
1 $2 \operatorname{acos} \frac{\phi}{2}$
2 $2 \operatorname{asin} \phi$
3 $2 \operatorname{acos} \phi$
4 $2 \operatorname{asin} \frac{\phi}{2}$
Explanation:
D Given, $\mathrm{y}_{1}=\mathrm{a} \sin (\mathrm{kx}-\omega \mathrm{t})$ and $\mathrm{y}_{2}=\mathrm{a} \sin (-\mathrm{k} x+\omega \mathrm{t}+\phi)$ Let, $\quad \mathrm{k} x-\omega \mathrm{t}=\alpha$ $\mathrm{y}_{1}=\mathrm{a} \sin \alpha, \mathrm{y}_{2}=\mathrm{a} \sin (\alpha-\phi)$ Superimposed wave, $\mathrm{y}=\mathrm{y}_{1}+\mathrm{y}_{2}$ $y=[a \sin \alpha+a \sin \phi \cos \alpha-a \cos \phi \sin \alpha]$ $y=[a(1-\cos \phi) \sin \alpha+(a \cos \phi) \cos \alpha]$ Let $\mathrm{R} \cos \theta=\mathrm{a}(1-\cos \phi)$ and $\mathrm{R} \sin \theta=\mathrm{a} \sin \phi$. Then, magnitude of $\mathrm{R}$ will be the magnitude of resultant wave, $\mathrm{R}^{2}=\mathrm{a}^{2}\left[\sin ^{2} \phi+(1-\cos \phi)^{2}\right]$ $\mathrm{R}^{2}=\mathrm{a}^{2}[1+1-2 \cos \phi]$ $\mathrm{R}^{2}=4 \mathrm{a}^{2} \sin ^{2} \frac{\phi}{2}$ Hence, $\quad \mathrm{R}=2 \mathrm{a} \sin \frac{\phi}{2}$
TS- EAMCET-04.05.2018
WAVES
172365
Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
D Resultant intensity of two periodic waves is given by $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2} \cos \phi}$ Where $\phi$ is the phase difference between two waves for maximum intensity $\phi=2 \pi \mathrm{n}(\mathrm{n}=0,1,2 \ldots \ldots \ldots)$ For maxima $\cos \phi=1$ $I_{\max }=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}}$ $I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$ $I_{\min } \text { when } \quad \cos 180^{\circ}=-1$ $I_{\text {min }}=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}$ $I_{\text {min }}=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$ So, the sum of the maximum and minimum intensity $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}}\right)^{2}+\left(\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}}\right)^{2}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}+\mathrm{I}_{1}+\mathrm{I}_{2}-2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=2\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)$
AIPMT-2008
WAVES
172366
Two waves represented by $y=a \sin (\omega t-k x)$ and $y=a \cos (\omega t-k x)$ are superimposed. The resultant wave will have an amplitude
172367
Three sinusoidal waves of the same frequency travel along a string in the positive $x$-direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, \pi / 2$ and $\pi$ respectively. What is the amplitude of the resultant wave?
172368
Two waves $y_{1}=2 \sin \omega t$ and $y_{2}=4 \sin (\omega t+\delta)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is
1 9
2 3
3 infinity
4 zero
Explanation:
A Given that, $\mathrm{y}_{1}=2 \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=4 \sin (\omega \mathrm{t}+\delta)$ Comparing the resultant wave equation, $\mathrm{y}_{1}=\mathrm{A}_{1} \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=\mathrm{A}_{2} \sin (\omega \mathrm{t}+\phi)$ Then, $\mathrm{A}_{1}=2, \mathrm{~A}_{2}=4$ So, $\quad \frac{I_{\text {max }}}{I_{\text {min }}}=\frac{\left(A_{1}+A_{2}\right)^{2}}{\left(A_{1}-A_{2}\right)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{(2+4)^{2}}{(2-4)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{36}{4}=9$
172364
Two harmonic travelling waves are described by the equations $y_{1}=a \sin (k x-\omega t)$ and $y_{2}=a$ $\sin (-k x+\omega t+\phi)$. The amplitude of the superimposed wave is
1 $2 \operatorname{acos} \frac{\phi}{2}$
2 $2 \operatorname{asin} \phi$
3 $2 \operatorname{acos} \phi$
4 $2 \operatorname{asin} \frac{\phi}{2}$
Explanation:
D Given, $\mathrm{y}_{1}=\mathrm{a} \sin (\mathrm{kx}-\omega \mathrm{t})$ and $\mathrm{y}_{2}=\mathrm{a} \sin (-\mathrm{k} x+\omega \mathrm{t}+\phi)$ Let, $\quad \mathrm{k} x-\omega \mathrm{t}=\alpha$ $\mathrm{y}_{1}=\mathrm{a} \sin \alpha, \mathrm{y}_{2}=\mathrm{a} \sin (\alpha-\phi)$ Superimposed wave, $\mathrm{y}=\mathrm{y}_{1}+\mathrm{y}_{2}$ $y=[a \sin \alpha+a \sin \phi \cos \alpha-a \cos \phi \sin \alpha]$ $y=[a(1-\cos \phi) \sin \alpha+(a \cos \phi) \cos \alpha]$ Let $\mathrm{R} \cos \theta=\mathrm{a}(1-\cos \phi)$ and $\mathrm{R} \sin \theta=\mathrm{a} \sin \phi$. Then, magnitude of $\mathrm{R}$ will be the magnitude of resultant wave, $\mathrm{R}^{2}=\mathrm{a}^{2}\left[\sin ^{2} \phi+(1-\cos \phi)^{2}\right]$ $\mathrm{R}^{2}=\mathrm{a}^{2}[1+1-2 \cos \phi]$ $\mathrm{R}^{2}=4 \mathrm{a}^{2} \sin ^{2} \frac{\phi}{2}$ Hence, $\quad \mathrm{R}=2 \mathrm{a} \sin \frac{\phi}{2}$
TS- EAMCET-04.05.2018
WAVES
172365
Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
D Resultant intensity of two periodic waves is given by $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2} \cos \phi}$ Where $\phi$ is the phase difference between two waves for maximum intensity $\phi=2 \pi \mathrm{n}(\mathrm{n}=0,1,2 \ldots \ldots \ldots)$ For maxima $\cos \phi=1$ $I_{\max }=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}}$ $I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$ $I_{\min } \text { when } \quad \cos 180^{\circ}=-1$ $I_{\text {min }}=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}$ $I_{\text {min }}=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$ So, the sum of the maximum and minimum intensity $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}}\right)^{2}+\left(\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}}\right)^{2}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}+\mathrm{I}_{1}+\mathrm{I}_{2}-2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=2\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)$
AIPMT-2008
WAVES
172366
Two waves represented by $y=a \sin (\omega t-k x)$ and $y=a \cos (\omega t-k x)$ are superimposed. The resultant wave will have an amplitude
172367
Three sinusoidal waves of the same frequency travel along a string in the positive $x$-direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, \pi / 2$ and $\pi$ respectively. What is the amplitude of the resultant wave?
172368
Two waves $y_{1}=2 \sin \omega t$ and $y_{2}=4 \sin (\omega t+\delta)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is
1 9
2 3
3 infinity
4 zero
Explanation:
A Given that, $\mathrm{y}_{1}=2 \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=4 \sin (\omega \mathrm{t}+\delta)$ Comparing the resultant wave equation, $\mathrm{y}_{1}=\mathrm{A}_{1} \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=\mathrm{A}_{2} \sin (\omega \mathrm{t}+\phi)$ Then, $\mathrm{A}_{1}=2, \mathrm{~A}_{2}=4$ So, $\quad \frac{I_{\text {max }}}{I_{\text {min }}}=\frac{\left(A_{1}+A_{2}\right)^{2}}{\left(A_{1}-A_{2}\right)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{(2+4)^{2}}{(2-4)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{36}{4}=9$
NEET Test Series from KOTA - 10 Papers In MS WORD
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WAVES
172364
Two harmonic travelling waves are described by the equations $y_{1}=a \sin (k x-\omega t)$ and $y_{2}=a$ $\sin (-k x+\omega t+\phi)$. The amplitude of the superimposed wave is
1 $2 \operatorname{acos} \frac{\phi}{2}$
2 $2 \operatorname{asin} \phi$
3 $2 \operatorname{acos} \phi$
4 $2 \operatorname{asin} \frac{\phi}{2}$
Explanation:
D Given, $\mathrm{y}_{1}=\mathrm{a} \sin (\mathrm{kx}-\omega \mathrm{t})$ and $\mathrm{y}_{2}=\mathrm{a} \sin (-\mathrm{k} x+\omega \mathrm{t}+\phi)$ Let, $\quad \mathrm{k} x-\omega \mathrm{t}=\alpha$ $\mathrm{y}_{1}=\mathrm{a} \sin \alpha, \mathrm{y}_{2}=\mathrm{a} \sin (\alpha-\phi)$ Superimposed wave, $\mathrm{y}=\mathrm{y}_{1}+\mathrm{y}_{2}$ $y=[a \sin \alpha+a \sin \phi \cos \alpha-a \cos \phi \sin \alpha]$ $y=[a(1-\cos \phi) \sin \alpha+(a \cos \phi) \cos \alpha]$ Let $\mathrm{R} \cos \theta=\mathrm{a}(1-\cos \phi)$ and $\mathrm{R} \sin \theta=\mathrm{a} \sin \phi$. Then, magnitude of $\mathrm{R}$ will be the magnitude of resultant wave, $\mathrm{R}^{2}=\mathrm{a}^{2}\left[\sin ^{2} \phi+(1-\cos \phi)^{2}\right]$ $\mathrm{R}^{2}=\mathrm{a}^{2}[1+1-2 \cos \phi]$ $\mathrm{R}^{2}=4 \mathrm{a}^{2} \sin ^{2} \frac{\phi}{2}$ Hence, $\quad \mathrm{R}=2 \mathrm{a} \sin \frac{\phi}{2}$
TS- EAMCET-04.05.2018
WAVES
172365
Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
D Resultant intensity of two periodic waves is given by $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2} \cos \phi}$ Where $\phi$ is the phase difference between two waves for maximum intensity $\phi=2 \pi \mathrm{n}(\mathrm{n}=0,1,2 \ldots \ldots \ldots)$ For maxima $\cos \phi=1$ $I_{\max }=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}}$ $I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$ $I_{\min } \text { when } \quad \cos 180^{\circ}=-1$ $I_{\text {min }}=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}$ $I_{\text {min }}=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$ So, the sum of the maximum and minimum intensity $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}}\right)^{2}+\left(\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}}\right)^{2}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}+\mathrm{I}_{1}+\mathrm{I}_{2}-2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=2\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)$
AIPMT-2008
WAVES
172366
Two waves represented by $y=a \sin (\omega t-k x)$ and $y=a \cos (\omega t-k x)$ are superimposed. The resultant wave will have an amplitude
172367
Three sinusoidal waves of the same frequency travel along a string in the positive $x$-direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, \pi / 2$ and $\pi$ respectively. What is the amplitude of the resultant wave?
172368
Two waves $y_{1}=2 \sin \omega t$ and $y_{2}=4 \sin (\omega t+\delta)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is
1 9
2 3
3 infinity
4 zero
Explanation:
A Given that, $\mathrm{y}_{1}=2 \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=4 \sin (\omega \mathrm{t}+\delta)$ Comparing the resultant wave equation, $\mathrm{y}_{1}=\mathrm{A}_{1} \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=\mathrm{A}_{2} \sin (\omega \mathrm{t}+\phi)$ Then, $\mathrm{A}_{1}=2, \mathrm{~A}_{2}=4$ So, $\quad \frac{I_{\text {max }}}{I_{\text {min }}}=\frac{\left(A_{1}+A_{2}\right)^{2}}{\left(A_{1}-A_{2}\right)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{(2+4)^{2}}{(2-4)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{36}{4}=9$
172364
Two harmonic travelling waves are described by the equations $y_{1}=a \sin (k x-\omega t)$ and $y_{2}=a$ $\sin (-k x+\omega t+\phi)$. The amplitude of the superimposed wave is
1 $2 \operatorname{acos} \frac{\phi}{2}$
2 $2 \operatorname{asin} \phi$
3 $2 \operatorname{acos} \phi$
4 $2 \operatorname{asin} \frac{\phi}{2}$
Explanation:
D Given, $\mathrm{y}_{1}=\mathrm{a} \sin (\mathrm{kx}-\omega \mathrm{t})$ and $\mathrm{y}_{2}=\mathrm{a} \sin (-\mathrm{k} x+\omega \mathrm{t}+\phi)$ Let, $\quad \mathrm{k} x-\omega \mathrm{t}=\alpha$ $\mathrm{y}_{1}=\mathrm{a} \sin \alpha, \mathrm{y}_{2}=\mathrm{a} \sin (\alpha-\phi)$ Superimposed wave, $\mathrm{y}=\mathrm{y}_{1}+\mathrm{y}_{2}$ $y=[a \sin \alpha+a \sin \phi \cos \alpha-a \cos \phi \sin \alpha]$ $y=[a(1-\cos \phi) \sin \alpha+(a \cos \phi) \cos \alpha]$ Let $\mathrm{R} \cos \theta=\mathrm{a}(1-\cos \phi)$ and $\mathrm{R} \sin \theta=\mathrm{a} \sin \phi$. Then, magnitude of $\mathrm{R}$ will be the magnitude of resultant wave, $\mathrm{R}^{2}=\mathrm{a}^{2}\left[\sin ^{2} \phi+(1-\cos \phi)^{2}\right]$ $\mathrm{R}^{2}=\mathrm{a}^{2}[1+1-2 \cos \phi]$ $\mathrm{R}^{2}=4 \mathrm{a}^{2} \sin ^{2} \frac{\phi}{2}$ Hence, $\quad \mathrm{R}=2 \mathrm{a} \sin \frac{\phi}{2}$
TS- EAMCET-04.05.2018
WAVES
172365
Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
D Resultant intensity of two periodic waves is given by $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2} \cos \phi}$ Where $\phi$ is the phase difference between two waves for maximum intensity $\phi=2 \pi \mathrm{n}(\mathrm{n}=0,1,2 \ldots \ldots \ldots)$ For maxima $\cos \phi=1$ $I_{\max }=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}}$ $I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$ $I_{\min } \text { when } \quad \cos 180^{\circ}=-1$ $I_{\text {min }}=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}$ $I_{\text {min }}=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$ So, the sum of the maximum and minimum intensity $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}}\right)^{2}+\left(\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}}\right)^{2}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}+\mathrm{I}_{1}+\mathrm{I}_{2}-2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=2\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)$
AIPMT-2008
WAVES
172366
Two waves represented by $y=a \sin (\omega t-k x)$ and $y=a \cos (\omega t-k x)$ are superimposed. The resultant wave will have an amplitude
172367
Three sinusoidal waves of the same frequency travel along a string in the positive $x$-direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, \pi / 2$ and $\pi$ respectively. What is the amplitude of the resultant wave?
172368
Two waves $y_{1}=2 \sin \omega t$ and $y_{2}=4 \sin (\omega t+\delta)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is
1 9
2 3
3 infinity
4 zero
Explanation:
A Given that, $\mathrm{y}_{1}=2 \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=4 \sin (\omega \mathrm{t}+\delta)$ Comparing the resultant wave equation, $\mathrm{y}_{1}=\mathrm{A}_{1} \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=\mathrm{A}_{2} \sin (\omega \mathrm{t}+\phi)$ Then, $\mathrm{A}_{1}=2, \mathrm{~A}_{2}=4$ So, $\quad \frac{I_{\text {max }}}{I_{\text {min }}}=\frac{\left(A_{1}+A_{2}\right)^{2}}{\left(A_{1}-A_{2}\right)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{(2+4)^{2}}{(2-4)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{36}{4}=9$
172364
Two harmonic travelling waves are described by the equations $y_{1}=a \sin (k x-\omega t)$ and $y_{2}=a$ $\sin (-k x+\omega t+\phi)$. The amplitude of the superimposed wave is
1 $2 \operatorname{acos} \frac{\phi}{2}$
2 $2 \operatorname{asin} \phi$
3 $2 \operatorname{acos} \phi$
4 $2 \operatorname{asin} \frac{\phi}{2}$
Explanation:
D Given, $\mathrm{y}_{1}=\mathrm{a} \sin (\mathrm{kx}-\omega \mathrm{t})$ and $\mathrm{y}_{2}=\mathrm{a} \sin (-\mathrm{k} x+\omega \mathrm{t}+\phi)$ Let, $\quad \mathrm{k} x-\omega \mathrm{t}=\alpha$ $\mathrm{y}_{1}=\mathrm{a} \sin \alpha, \mathrm{y}_{2}=\mathrm{a} \sin (\alpha-\phi)$ Superimposed wave, $\mathrm{y}=\mathrm{y}_{1}+\mathrm{y}_{2}$ $y=[a \sin \alpha+a \sin \phi \cos \alpha-a \cos \phi \sin \alpha]$ $y=[a(1-\cos \phi) \sin \alpha+(a \cos \phi) \cos \alpha]$ Let $\mathrm{R} \cos \theta=\mathrm{a}(1-\cos \phi)$ and $\mathrm{R} \sin \theta=\mathrm{a} \sin \phi$. Then, magnitude of $\mathrm{R}$ will be the magnitude of resultant wave, $\mathrm{R}^{2}=\mathrm{a}^{2}\left[\sin ^{2} \phi+(1-\cos \phi)^{2}\right]$ $\mathrm{R}^{2}=\mathrm{a}^{2}[1+1-2 \cos \phi]$ $\mathrm{R}^{2}=4 \mathrm{a}^{2} \sin ^{2} \frac{\phi}{2}$ Hence, $\quad \mathrm{R}=2 \mathrm{a} \sin \frac{\phi}{2}$
TS- EAMCET-04.05.2018
WAVES
172365
Two periodic waves of intensities $I_{1}$ and $I_{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is
D Resultant intensity of two periodic waves is given by $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2} \cos \phi}$ Where $\phi$ is the phase difference between two waves for maximum intensity $\phi=2 \pi \mathrm{n}(\mathrm{n}=0,1,2 \ldots \ldots \ldots)$ For maxima $\cos \phi=1$ $I_{\max }=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}}$ $I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}$ $I_{\min } \text { when } \quad \cos 180^{\circ}=-1$ $I_{\text {min }}=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}$ $I_{\text {min }}=\left(\sqrt{I_{1}}-\sqrt{I_{2}}\right)^{2}$ So, the sum of the maximum and minimum intensity $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}}\right)^{2}+\left(\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}}\right)^{2}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}+\mathrm{I}_{1}+\mathrm{I}_{2}-2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}$ $\mathrm{I}_{\max }+\mathrm{I}_{\min }=2\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right)$
AIPMT-2008
WAVES
172366
Two waves represented by $y=a \sin (\omega t-k x)$ and $y=a \cos (\omega t-k x)$ are superimposed. The resultant wave will have an amplitude
172367
Three sinusoidal waves of the same frequency travel along a string in the positive $x$-direction. Their amplitudes are $y, y / 2$ and $y / 3$ and their phase constants are $0, \pi / 2$ and $\pi$ respectively. What is the amplitude of the resultant wave?
172368
Two waves $y_{1}=2 \sin \omega t$ and $y_{2}=4 \sin (\omega t+\delta)$ superimpose. The ratio of the maximum to the minimum intensity of the resultant wave is
1 9
2 3
3 infinity
4 zero
Explanation:
A Given that, $\mathrm{y}_{1}=2 \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=4 \sin (\omega \mathrm{t}+\delta)$ Comparing the resultant wave equation, $\mathrm{y}_{1}=\mathrm{A}_{1} \sin \omega \mathrm{t}$ $\mathrm{y}_{2}=\mathrm{A}_{2} \sin (\omega \mathrm{t}+\phi)$ Then, $\mathrm{A}_{1}=2, \mathrm{~A}_{2}=4$ So, $\quad \frac{I_{\text {max }}}{I_{\text {min }}}=\frac{\left(A_{1}+A_{2}\right)^{2}}{\left(A_{1}-A_{2}\right)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{(2+4)^{2}}{(2-4)^{2}}$ $\frac{I_{\max }}{I_{\min }}=\frac{36}{4}=9$