172338
Equations of motion in the same direction are given by, $\mathrm{y}_{1}=2 \mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{y}_{2}=2 \mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}-\theta)$ The amplitude of the medium particle will be
172339
A progressive wave moving along $x$-axis is represented by $y=A \sin \left[\frac{2 \pi}{\lambda}(v t-x)\right]$. The wavelength $(\lambda)$ at which the maximum particle velocity is 3 times the wave velocity is
172345
The equation of wave is $y=1.0 \cos 2 \pi\left(\frac{t}{0.02}-\frac{x}{10}\right)$ where $t$ is in second. The frequency of the wave is
1 $50 \mathrm{~Hz}$
2 $315 \mathrm{~Hz}$
3 $10 \mathrm{~Hz}$
4 $63 \mathrm{~Hz}$
Explanation:
A Given, $y=1.0 \cos 2 \pi\left(\frac{t}{0.02}-\frac{x}{10}\right)$ $y=1.0 \cos \left(\frac{2 \pi t}{0.02}-\frac{2 \pi x}{10}\right)$ $y=A \cos [\omega t-k x]$ Comparing equation (i) and (ii), we get - $\omega=\frac{2 \pi}{0.02}$ We know that, $\omega=2 \pi \mathrm{f}$ $2 \pi \mathrm{f} =\frac{2 \pi}{0.02}$ $\mathrm{f} =\frac{1}{0.02}=\frac{100}{2}=50 \mathrm{~Hz}$
AP EAMCET - 1998
WAVES
172347
What is the path difference between the waves $y_{1}=a \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ and $y_{2}=a \cos \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ ?
1 $\lambda$
2 $\frac{\lambda}{2}$
3 $\frac{\lambda}{4}$
4 $2 \lambda$
Explanation:
C Given that displacement equation - $\mathrm{y}_{1}=\mathrm{a} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ $\mathrm{y}_{2}=\mathrm{a} \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ So, $\mathrm{y}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\frac{\pi}{2}\right)$ Then phase different is $\frac{\pi}{2}$ Thus path difference, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times$ phase difference $\Delta x =\frac{\lambda}{2 \pi} \times \frac{\pi}{2}$ $\Delta x =\frac{\lambda}{4}$
172338
Equations of motion in the same direction are given by, $\mathrm{y}_{1}=2 \mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{y}_{2}=2 \mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}-\theta)$ The amplitude of the medium particle will be
172339
A progressive wave moving along $x$-axis is represented by $y=A \sin \left[\frac{2 \pi}{\lambda}(v t-x)\right]$. The wavelength $(\lambda)$ at which the maximum particle velocity is 3 times the wave velocity is
172345
The equation of wave is $y=1.0 \cos 2 \pi\left(\frac{t}{0.02}-\frac{x}{10}\right)$ where $t$ is in second. The frequency of the wave is
1 $50 \mathrm{~Hz}$
2 $315 \mathrm{~Hz}$
3 $10 \mathrm{~Hz}$
4 $63 \mathrm{~Hz}$
Explanation:
A Given, $y=1.0 \cos 2 \pi\left(\frac{t}{0.02}-\frac{x}{10}\right)$ $y=1.0 \cos \left(\frac{2 \pi t}{0.02}-\frac{2 \pi x}{10}\right)$ $y=A \cos [\omega t-k x]$ Comparing equation (i) and (ii), we get - $\omega=\frac{2 \pi}{0.02}$ We know that, $\omega=2 \pi \mathrm{f}$ $2 \pi \mathrm{f} =\frac{2 \pi}{0.02}$ $\mathrm{f} =\frac{1}{0.02}=\frac{100}{2}=50 \mathrm{~Hz}$
AP EAMCET - 1998
WAVES
172347
What is the path difference between the waves $y_{1}=a \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ and $y_{2}=a \cos \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ ?
1 $\lambda$
2 $\frac{\lambda}{2}$
3 $\frac{\lambda}{4}$
4 $2 \lambda$
Explanation:
C Given that displacement equation - $\mathrm{y}_{1}=\mathrm{a} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ $\mathrm{y}_{2}=\mathrm{a} \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ So, $\mathrm{y}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\frac{\pi}{2}\right)$ Then phase different is $\frac{\pi}{2}$ Thus path difference, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times$ phase difference $\Delta x =\frac{\lambda}{2 \pi} \times \frac{\pi}{2}$ $\Delta x =\frac{\lambda}{4}$
172338
Equations of motion in the same direction are given by, $\mathrm{y}_{1}=2 \mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{y}_{2}=2 \mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}-\theta)$ The amplitude of the medium particle will be
172339
A progressive wave moving along $x$-axis is represented by $y=A \sin \left[\frac{2 \pi}{\lambda}(v t-x)\right]$. The wavelength $(\lambda)$ at which the maximum particle velocity is 3 times the wave velocity is
172345
The equation of wave is $y=1.0 \cos 2 \pi\left(\frac{t}{0.02}-\frac{x}{10}\right)$ where $t$ is in second. The frequency of the wave is
1 $50 \mathrm{~Hz}$
2 $315 \mathrm{~Hz}$
3 $10 \mathrm{~Hz}$
4 $63 \mathrm{~Hz}$
Explanation:
A Given, $y=1.0 \cos 2 \pi\left(\frac{t}{0.02}-\frac{x}{10}\right)$ $y=1.0 \cos \left(\frac{2 \pi t}{0.02}-\frac{2 \pi x}{10}\right)$ $y=A \cos [\omega t-k x]$ Comparing equation (i) and (ii), we get - $\omega=\frac{2 \pi}{0.02}$ We know that, $\omega=2 \pi \mathrm{f}$ $2 \pi \mathrm{f} =\frac{2 \pi}{0.02}$ $\mathrm{f} =\frac{1}{0.02}=\frac{100}{2}=50 \mathrm{~Hz}$
AP EAMCET - 1998
WAVES
172347
What is the path difference between the waves $y_{1}=a \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ and $y_{2}=a \cos \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ ?
1 $\lambda$
2 $\frac{\lambda}{2}$
3 $\frac{\lambda}{4}$
4 $2 \lambda$
Explanation:
C Given that displacement equation - $\mathrm{y}_{1}=\mathrm{a} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ $\mathrm{y}_{2}=\mathrm{a} \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ So, $\mathrm{y}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\frac{\pi}{2}\right)$ Then phase different is $\frac{\pi}{2}$ Thus path difference, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times$ phase difference $\Delta x =\frac{\lambda}{2 \pi} \times \frac{\pi}{2}$ $\Delta x =\frac{\lambda}{4}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
WAVES
172338
Equations of motion in the same direction are given by, $\mathrm{y}_{1}=2 \mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})$ $\mathrm{y}_{2}=2 \mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx}-\theta)$ The amplitude of the medium particle will be
172339
A progressive wave moving along $x$-axis is represented by $y=A \sin \left[\frac{2 \pi}{\lambda}(v t-x)\right]$. The wavelength $(\lambda)$ at which the maximum particle velocity is 3 times the wave velocity is
172345
The equation of wave is $y=1.0 \cos 2 \pi\left(\frac{t}{0.02}-\frac{x}{10}\right)$ where $t$ is in second. The frequency of the wave is
1 $50 \mathrm{~Hz}$
2 $315 \mathrm{~Hz}$
3 $10 \mathrm{~Hz}$
4 $63 \mathrm{~Hz}$
Explanation:
A Given, $y=1.0 \cos 2 \pi\left(\frac{t}{0.02}-\frac{x}{10}\right)$ $y=1.0 \cos \left(\frac{2 \pi t}{0.02}-\frac{2 \pi x}{10}\right)$ $y=A \cos [\omega t-k x]$ Comparing equation (i) and (ii), we get - $\omega=\frac{2 \pi}{0.02}$ We know that, $\omega=2 \pi \mathrm{f}$ $2 \pi \mathrm{f} =\frac{2 \pi}{0.02}$ $\mathrm{f} =\frac{1}{0.02}=\frac{100}{2}=50 \mathrm{~Hz}$
AP EAMCET - 1998
WAVES
172347
What is the path difference between the waves $y_{1}=a \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ and $y_{2}=a \cos \left(\omega t-\frac{2 \pi x}{\lambda}\right)$ ?
1 $\lambda$
2 $\frac{\lambda}{2}$
3 $\frac{\lambda}{4}$
4 $2 \lambda$
Explanation:
C Given that displacement equation - $\mathrm{y}_{1}=\mathrm{a} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ $\mathrm{y}_{2}=\mathrm{a} \cos \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}\right)$ So, $\mathrm{y}_{2}=\mathrm{a} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\frac{\pi}{2}\right)$ Then phase different is $\frac{\pi}{2}$ Thus path difference, $\Delta \mathrm{x}=\frac{\lambda}{2 \pi} \times$ phase difference $\Delta x =\frac{\lambda}{2 \pi} \times \frac{\pi}{2}$ $\Delta x =\frac{\lambda}{4}$
