172221
The phase difference between two points is $\pi / 3$. If the frequency of wave is $50 \mathrm{~Hz}$, then what is the distance between two points? (Given, $v=330 \mathrm{~ms}^{1}$ )
172224
An observer standing near the sea shore observes 54 waves per minute. If the wavelength of the water wave is $10 \mathrm{~m}$ then the velocity of water wave is:
1 $540 \mathrm{~ms}^{-1}$
2 $5.4 \mathrm{~ms}^{-1}$
3 $0.184 \mathrm{~ms}^{-1}$
4 $9 \mathrm{~ms}^{-1}$
5 $48.6 \mathrm{~ms}^{-1}$
Explanation:
D Given $\lambda=10 \mathrm{~m}$, No. of wave per minute $=54$ So, No. of wave per second $=\frac{54}{60} \mathrm{~Hz}$. Now, $\text { Velocity, } \mathrm{v}=\mathrm{f} \lambda$ $=\frac{54}{60} \times 10$ $=9 \mathrm{~m} / \mathrm{sec}$
Kerala CEE 2005
WAVES
172225
The equation $y=A \sin 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$, where the symbols carry the usual meaning and $A, T$ and $\lambda$ are positive, represents a wave of:
172227
The equation of a travelling wave is given by $y=0.5 \sin (20 x-400 t)$ where $x$ and $y$ are in meter and $t$ is in second. The velocity of the wave is :
1 $10 \mathrm{~m} / \mathrm{s}$
2 $20 \mathrm{~m} / \mathrm{s}$
3 $200 \mathrm{~m} / \mathrm{s}$
4 $400 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, $y=0.5 \sin (20 x-400 t)$ Comparing on general equation, $y=a \sin (k x-\omega t)$ We get $\omega=400$ $k=20$ Velocity of wave, $\mathrm{v} =\frac{\omega}{\mathrm{k}}$ $\mathrm{v} =\frac{400}{20}$ $\mathrm{v} =20 \mathrm{~m} / \mathrm{s}$
172221
The phase difference between two points is $\pi / 3$. If the frequency of wave is $50 \mathrm{~Hz}$, then what is the distance between two points? (Given, $v=330 \mathrm{~ms}^{1}$ )
172224
An observer standing near the sea shore observes 54 waves per minute. If the wavelength of the water wave is $10 \mathrm{~m}$ then the velocity of water wave is:
1 $540 \mathrm{~ms}^{-1}$
2 $5.4 \mathrm{~ms}^{-1}$
3 $0.184 \mathrm{~ms}^{-1}$
4 $9 \mathrm{~ms}^{-1}$
5 $48.6 \mathrm{~ms}^{-1}$
Explanation:
D Given $\lambda=10 \mathrm{~m}$, No. of wave per minute $=54$ So, No. of wave per second $=\frac{54}{60} \mathrm{~Hz}$. Now, $\text { Velocity, } \mathrm{v}=\mathrm{f} \lambda$ $=\frac{54}{60} \times 10$ $=9 \mathrm{~m} / \mathrm{sec}$
Kerala CEE 2005
WAVES
172225
The equation $y=A \sin 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$, where the symbols carry the usual meaning and $A, T$ and $\lambda$ are positive, represents a wave of:
172227
The equation of a travelling wave is given by $y=0.5 \sin (20 x-400 t)$ where $x$ and $y$ are in meter and $t$ is in second. The velocity of the wave is :
1 $10 \mathrm{~m} / \mathrm{s}$
2 $20 \mathrm{~m} / \mathrm{s}$
3 $200 \mathrm{~m} / \mathrm{s}$
4 $400 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, $y=0.5 \sin (20 x-400 t)$ Comparing on general equation, $y=a \sin (k x-\omega t)$ We get $\omega=400$ $k=20$ Velocity of wave, $\mathrm{v} =\frac{\omega}{\mathrm{k}}$ $\mathrm{v} =\frac{400}{20}$ $\mathrm{v} =20 \mathrm{~m} / \mathrm{s}$
172221
The phase difference between two points is $\pi / 3$. If the frequency of wave is $50 \mathrm{~Hz}$, then what is the distance between two points? (Given, $v=330 \mathrm{~ms}^{1}$ )
172224
An observer standing near the sea shore observes 54 waves per minute. If the wavelength of the water wave is $10 \mathrm{~m}$ then the velocity of water wave is:
1 $540 \mathrm{~ms}^{-1}$
2 $5.4 \mathrm{~ms}^{-1}$
3 $0.184 \mathrm{~ms}^{-1}$
4 $9 \mathrm{~ms}^{-1}$
5 $48.6 \mathrm{~ms}^{-1}$
Explanation:
D Given $\lambda=10 \mathrm{~m}$, No. of wave per minute $=54$ So, No. of wave per second $=\frac{54}{60} \mathrm{~Hz}$. Now, $\text { Velocity, } \mathrm{v}=\mathrm{f} \lambda$ $=\frac{54}{60} \times 10$ $=9 \mathrm{~m} / \mathrm{sec}$
Kerala CEE 2005
WAVES
172225
The equation $y=A \sin 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$, where the symbols carry the usual meaning and $A, T$ and $\lambda$ are positive, represents a wave of:
172227
The equation of a travelling wave is given by $y=0.5 \sin (20 x-400 t)$ where $x$ and $y$ are in meter and $t$ is in second. The velocity of the wave is :
1 $10 \mathrm{~m} / \mathrm{s}$
2 $20 \mathrm{~m} / \mathrm{s}$
3 $200 \mathrm{~m} / \mathrm{s}$
4 $400 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, $y=0.5 \sin (20 x-400 t)$ Comparing on general equation, $y=a \sin (k x-\omega t)$ We get $\omega=400$ $k=20$ Velocity of wave, $\mathrm{v} =\frac{\omega}{\mathrm{k}}$ $\mathrm{v} =\frac{400}{20}$ $\mathrm{v} =20 \mathrm{~m} / \mathrm{s}$
172221
The phase difference between two points is $\pi / 3$. If the frequency of wave is $50 \mathrm{~Hz}$, then what is the distance between two points? (Given, $v=330 \mathrm{~ms}^{1}$ )
172224
An observer standing near the sea shore observes 54 waves per minute. If the wavelength of the water wave is $10 \mathrm{~m}$ then the velocity of water wave is:
1 $540 \mathrm{~ms}^{-1}$
2 $5.4 \mathrm{~ms}^{-1}$
3 $0.184 \mathrm{~ms}^{-1}$
4 $9 \mathrm{~ms}^{-1}$
5 $48.6 \mathrm{~ms}^{-1}$
Explanation:
D Given $\lambda=10 \mathrm{~m}$, No. of wave per minute $=54$ So, No. of wave per second $=\frac{54}{60} \mathrm{~Hz}$. Now, $\text { Velocity, } \mathrm{v}=\mathrm{f} \lambda$ $=\frac{54}{60} \times 10$ $=9 \mathrm{~m} / \mathrm{sec}$
Kerala CEE 2005
WAVES
172225
The equation $y=A \sin 2 \pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)$, where the symbols carry the usual meaning and $A, T$ and $\lambda$ are positive, represents a wave of:
172227
The equation of a travelling wave is given by $y=0.5 \sin (20 x-400 t)$ where $x$ and $y$ are in meter and $t$ is in second. The velocity of the wave is :
1 $10 \mathrm{~m} / \mathrm{s}$
2 $20 \mathrm{~m} / \mathrm{s}$
3 $200 \mathrm{~m} / \mathrm{s}$
4 $400 \mathrm{~m} / \mathrm{s}$
Explanation:
B Given, $y=0.5 \sin (20 x-400 t)$ Comparing on general equation, $y=a \sin (k x-\omega t)$ We get $\omega=400$ $k=20$ Velocity of wave, $\mathrm{v} =\frac{\omega}{\mathrm{k}}$ $\mathrm{v} =\frac{400}{20}$ $\mathrm{v} =20 \mathrm{~m} / \mathrm{s}$
