172253
The wave equation is $y=0.30 \sin (314 t-1.57$ $x)$ where $t, x$ and $y$ are in second, meter and centimeter respectively. The speed of the wave is
1 $400 \mathrm{~m} / \mathrm{s}$
2 $100 \mathrm{~m} / \mathrm{s}$
3 $200 \mathrm{~m} / \mathrm{s}$
4 $50 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, $\mathrm{y}=0.30 \sin (314 \mathrm{t}-1.57 \mathrm{x})$ Compared by, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Then, $\omega=314$ and $\mathrm{k}=1.57$ Speed of a wave, $v=\frac{\omega}{\mathrm{k}}$ $\mathrm{v}=\frac{314}{1.57}$ $\mathrm{v}=200 \mathrm{~m} / \mathrm{s}$
BCECE-2004
WAVES
172254
The displacement of a travelling wave is given by- $y=2 \cos 2 \pi(10 t-0.008 x+0.35)$ where, $x$ and $y$ are in centimetre and $t$ in second. What is the phase difference between oscillatory motion at two points separated by a distance of $4 \mathbf{m}$ ?
1 $2 \pi$
2 $4 \pi$
3 $6 \pi$
4 $8 \pi$
Explanation:
C Given, Path difference, $\Delta \mathrm{x}=4 \mathrm{~m}$ The equation of travelling wave is, $y=2 \cos 2 \pi(10 t-0.008 x+0.35)$ We can write as the above equation $\mathrm{y}=2 \cos (20 \pi \mathrm{t}-0.016 \pi \mathrm{x}+0.70 \pi)$ Compare the above equation with standard equation of wave, we get $\omega=20 \pi, \mathrm{a}=2 \mathrm{~cm}, \mathrm{k}=0.016 \pi$ $\mathrm{k}=\frac{2 \pi}{\lambda}$ $\lambda=\frac{2 \times \pi}{0.016 \pi}$ $\lambda=\frac{1}{0.008}$ $\lambda=125 \mathrm{~cm}=1.25 \mathrm{~m}$ Phase difference $(\Delta \phi)=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ $=\frac{2 \times \pi}{1.25} \times 4$ $\Delta \phi \square 6.4 \pi$ $\Delta \phi =6 \pi$
BCECE-2009
WAVES
172256
A wave travelling in the positive $X$-direction having maximum displacement along $\mathrm{Y}$ direction as $1 \mathrm{~m}$, wavelength $2 \pi \mathrm{m}$ and frequency of $\frac{1}{\pi} \mathrm{Hz}$ is represented by
A Given that, $\mathrm{A}=1 \mathrm{~m}, \lambda=2 \pi \mathrm{m}, \mathrm{f}=\frac{1}{\pi}$ $\omega=2 \pi \mathrm{f}$ $\omega=2 \pi \times \frac{1}{\pi}=2$ $\mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi}{2 \pi}=1$ From the standard equation of wave, $\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t})$ After putting value of $\mathrm{A}, \omega$ and $\mathrm{k}$ $y=\sin (x-2 t)$
NEET-2013
WAVES
172257
The equation of sound wave is $y=0.0015 \sin (62.4 x+316 t)$. Find the wavelength of this wave
1 0.2 unit
2 0.1 unit
3 0.3 unit
4 None of these
Explanation:
B Given, $y=0.0015 \sin (62.4 x+316 t)$ Comparing it with the equation $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\mathrm{kx})$, we get $\omega =316, \mathrm{k}=62.4$ $\mathrm{k} =\frac{2 \pi}{\lambda}=62.4$ $\lambda =\frac{2 \pi}{62.4}=0.1$ $\lambda =0.1 \text { unit }$
172253
The wave equation is $y=0.30 \sin (314 t-1.57$ $x)$ where $t, x$ and $y$ are in second, meter and centimeter respectively. The speed of the wave is
1 $400 \mathrm{~m} / \mathrm{s}$
2 $100 \mathrm{~m} / \mathrm{s}$
3 $200 \mathrm{~m} / \mathrm{s}$
4 $50 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, $\mathrm{y}=0.30 \sin (314 \mathrm{t}-1.57 \mathrm{x})$ Compared by, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Then, $\omega=314$ and $\mathrm{k}=1.57$ Speed of a wave, $v=\frac{\omega}{\mathrm{k}}$ $\mathrm{v}=\frac{314}{1.57}$ $\mathrm{v}=200 \mathrm{~m} / \mathrm{s}$
BCECE-2004
WAVES
172254
The displacement of a travelling wave is given by- $y=2 \cos 2 \pi(10 t-0.008 x+0.35)$ where, $x$ and $y$ are in centimetre and $t$ in second. What is the phase difference between oscillatory motion at two points separated by a distance of $4 \mathbf{m}$ ?
1 $2 \pi$
2 $4 \pi$
3 $6 \pi$
4 $8 \pi$
Explanation:
C Given, Path difference, $\Delta \mathrm{x}=4 \mathrm{~m}$ The equation of travelling wave is, $y=2 \cos 2 \pi(10 t-0.008 x+0.35)$ We can write as the above equation $\mathrm{y}=2 \cos (20 \pi \mathrm{t}-0.016 \pi \mathrm{x}+0.70 \pi)$ Compare the above equation with standard equation of wave, we get $\omega=20 \pi, \mathrm{a}=2 \mathrm{~cm}, \mathrm{k}=0.016 \pi$ $\mathrm{k}=\frac{2 \pi}{\lambda}$ $\lambda=\frac{2 \times \pi}{0.016 \pi}$ $\lambda=\frac{1}{0.008}$ $\lambda=125 \mathrm{~cm}=1.25 \mathrm{~m}$ Phase difference $(\Delta \phi)=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ $=\frac{2 \times \pi}{1.25} \times 4$ $\Delta \phi \square 6.4 \pi$ $\Delta \phi =6 \pi$
BCECE-2009
WAVES
172256
A wave travelling in the positive $X$-direction having maximum displacement along $\mathrm{Y}$ direction as $1 \mathrm{~m}$, wavelength $2 \pi \mathrm{m}$ and frequency of $\frac{1}{\pi} \mathrm{Hz}$ is represented by
A Given that, $\mathrm{A}=1 \mathrm{~m}, \lambda=2 \pi \mathrm{m}, \mathrm{f}=\frac{1}{\pi}$ $\omega=2 \pi \mathrm{f}$ $\omega=2 \pi \times \frac{1}{\pi}=2$ $\mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi}{2 \pi}=1$ From the standard equation of wave, $\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t})$ After putting value of $\mathrm{A}, \omega$ and $\mathrm{k}$ $y=\sin (x-2 t)$
NEET-2013
WAVES
172257
The equation of sound wave is $y=0.0015 \sin (62.4 x+316 t)$. Find the wavelength of this wave
1 0.2 unit
2 0.1 unit
3 0.3 unit
4 None of these
Explanation:
B Given, $y=0.0015 \sin (62.4 x+316 t)$ Comparing it with the equation $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\mathrm{kx})$, we get $\omega =316, \mathrm{k}=62.4$ $\mathrm{k} =\frac{2 \pi}{\lambda}=62.4$ $\lambda =\frac{2 \pi}{62.4}=0.1$ $\lambda =0.1 \text { unit }$
172253
The wave equation is $y=0.30 \sin (314 t-1.57$ $x)$ where $t, x$ and $y$ are in second, meter and centimeter respectively. The speed of the wave is
1 $400 \mathrm{~m} / \mathrm{s}$
2 $100 \mathrm{~m} / \mathrm{s}$
3 $200 \mathrm{~m} / \mathrm{s}$
4 $50 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, $\mathrm{y}=0.30 \sin (314 \mathrm{t}-1.57 \mathrm{x})$ Compared by, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Then, $\omega=314$ and $\mathrm{k}=1.57$ Speed of a wave, $v=\frac{\omega}{\mathrm{k}}$ $\mathrm{v}=\frac{314}{1.57}$ $\mathrm{v}=200 \mathrm{~m} / \mathrm{s}$
BCECE-2004
WAVES
172254
The displacement of a travelling wave is given by- $y=2 \cos 2 \pi(10 t-0.008 x+0.35)$ where, $x$ and $y$ are in centimetre and $t$ in second. What is the phase difference between oscillatory motion at two points separated by a distance of $4 \mathbf{m}$ ?
1 $2 \pi$
2 $4 \pi$
3 $6 \pi$
4 $8 \pi$
Explanation:
C Given, Path difference, $\Delta \mathrm{x}=4 \mathrm{~m}$ The equation of travelling wave is, $y=2 \cos 2 \pi(10 t-0.008 x+0.35)$ We can write as the above equation $\mathrm{y}=2 \cos (20 \pi \mathrm{t}-0.016 \pi \mathrm{x}+0.70 \pi)$ Compare the above equation with standard equation of wave, we get $\omega=20 \pi, \mathrm{a}=2 \mathrm{~cm}, \mathrm{k}=0.016 \pi$ $\mathrm{k}=\frac{2 \pi}{\lambda}$ $\lambda=\frac{2 \times \pi}{0.016 \pi}$ $\lambda=\frac{1}{0.008}$ $\lambda=125 \mathrm{~cm}=1.25 \mathrm{~m}$ Phase difference $(\Delta \phi)=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ $=\frac{2 \times \pi}{1.25} \times 4$ $\Delta \phi \square 6.4 \pi$ $\Delta \phi =6 \pi$
BCECE-2009
WAVES
172256
A wave travelling in the positive $X$-direction having maximum displacement along $\mathrm{Y}$ direction as $1 \mathrm{~m}$, wavelength $2 \pi \mathrm{m}$ and frequency of $\frac{1}{\pi} \mathrm{Hz}$ is represented by
A Given that, $\mathrm{A}=1 \mathrm{~m}, \lambda=2 \pi \mathrm{m}, \mathrm{f}=\frac{1}{\pi}$ $\omega=2 \pi \mathrm{f}$ $\omega=2 \pi \times \frac{1}{\pi}=2$ $\mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi}{2 \pi}=1$ From the standard equation of wave, $\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t})$ After putting value of $\mathrm{A}, \omega$ and $\mathrm{k}$ $y=\sin (x-2 t)$
NEET-2013
WAVES
172257
The equation of sound wave is $y=0.0015 \sin (62.4 x+316 t)$. Find the wavelength of this wave
1 0.2 unit
2 0.1 unit
3 0.3 unit
4 None of these
Explanation:
B Given, $y=0.0015 \sin (62.4 x+316 t)$ Comparing it with the equation $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\mathrm{kx})$, we get $\omega =316, \mathrm{k}=62.4$ $\mathrm{k} =\frac{2 \pi}{\lambda}=62.4$ $\lambda =\frac{2 \pi}{62.4}=0.1$ $\lambda =0.1 \text { unit }$
172253
The wave equation is $y=0.30 \sin (314 t-1.57$ $x)$ where $t, x$ and $y$ are in second, meter and centimeter respectively. The speed of the wave is
1 $400 \mathrm{~m} / \mathrm{s}$
2 $100 \mathrm{~m} / \mathrm{s}$
3 $200 \mathrm{~m} / \mathrm{s}$
4 $50 \mathrm{~m} / \mathrm{s}$
Explanation:
C Given, $\mathrm{y}=0.30 \sin (314 \mathrm{t}-1.57 \mathrm{x})$ Compared by, $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}-\mathrm{kx})$ Then, $\omega=314$ and $\mathrm{k}=1.57$ Speed of a wave, $v=\frac{\omega}{\mathrm{k}}$ $\mathrm{v}=\frac{314}{1.57}$ $\mathrm{v}=200 \mathrm{~m} / \mathrm{s}$
BCECE-2004
WAVES
172254
The displacement of a travelling wave is given by- $y=2 \cos 2 \pi(10 t-0.008 x+0.35)$ where, $x$ and $y$ are in centimetre and $t$ in second. What is the phase difference between oscillatory motion at two points separated by a distance of $4 \mathbf{m}$ ?
1 $2 \pi$
2 $4 \pi$
3 $6 \pi$
4 $8 \pi$
Explanation:
C Given, Path difference, $\Delta \mathrm{x}=4 \mathrm{~m}$ The equation of travelling wave is, $y=2 \cos 2 \pi(10 t-0.008 x+0.35)$ We can write as the above equation $\mathrm{y}=2 \cos (20 \pi \mathrm{t}-0.016 \pi \mathrm{x}+0.70 \pi)$ Compare the above equation with standard equation of wave, we get $\omega=20 \pi, \mathrm{a}=2 \mathrm{~cm}, \mathrm{k}=0.016 \pi$ $\mathrm{k}=\frac{2 \pi}{\lambda}$ $\lambda=\frac{2 \times \pi}{0.016 \pi}$ $\lambda=\frac{1}{0.008}$ $\lambda=125 \mathrm{~cm}=1.25 \mathrm{~m}$ Phase difference $(\Delta \phi)=\frac{2 \pi}{\lambda} \times \Delta \mathrm{x}$ $=\frac{2 \times \pi}{1.25} \times 4$ $\Delta \phi \square 6.4 \pi$ $\Delta \phi =6 \pi$
BCECE-2009
WAVES
172256
A wave travelling in the positive $X$-direction having maximum displacement along $\mathrm{Y}$ direction as $1 \mathrm{~m}$, wavelength $2 \pi \mathrm{m}$ and frequency of $\frac{1}{\pi} \mathrm{Hz}$ is represented by
A Given that, $\mathrm{A}=1 \mathrm{~m}, \lambda=2 \pi \mathrm{m}, \mathrm{f}=\frac{1}{\pi}$ $\omega=2 \pi \mathrm{f}$ $\omega=2 \pi \times \frac{1}{\pi}=2$ $\mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi}{2 \pi}=1$ From the standard equation of wave, $\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}-\omega \mathrm{t})$ After putting value of $\mathrm{A}, \omega$ and $\mathrm{k}$ $y=\sin (x-2 t)$
NEET-2013
WAVES
172257
The equation of sound wave is $y=0.0015 \sin (62.4 x+316 t)$. Find the wavelength of this wave
1 0.2 unit
2 0.1 unit
3 0.3 unit
4 None of these
Explanation:
B Given, $y=0.0015 \sin (62.4 x+316 t)$ Comparing it with the equation $\mathrm{y}=\mathrm{A} \sin (\omega \mathrm{t}+\mathrm{kx})$, we get $\omega =316, \mathrm{k}=62.4$ $\mathrm{k} =\frac{2 \pi}{\lambda}=62.4$ $\lambda =\frac{2 \pi}{62.4}=0.1$ $\lambda =0.1 \text { unit }$
