Answer
409.5k+ views
Hint: Here we have to first find the equation given by the displacement function and compare the given equation and the displacement equation to get the speed. Vibration of a wave is represented by a transverse harmonic wave.
Then we have to find the frequency of the given wave by using the formula between frequency and angular velocity.
Also we can get the phase angle by comparing the displacement equation and the given equation.
Lastly, we can find the wavelength by determining the distance between the successive crests.
Complete step by step solution:
(A). The displacement equation of wave travelling from left to right is given by:
$y(x,t) = a\sin (\omega t + kx + \phi )$
...... (1)
The given equation in the question is:
$y(x,t) = 3.0\sin (36t + 0.018x + \dfrac{\pi }{4})$
...... (2)
When we compare the two equations, we find that equation (2) also represents the displacement equation of a wave. So, it is a travelling wave.
From equations (1) and (2) we get,
$\omega = 36\,rad/s$, $k = 0.018\,{m^{ - 1}}$
We know that,
Speed, $v = \dfrac{\omega }{{2\pi }}$ and wavelength, $\lambda = \dfrac{{2\pi }}{k}$
Also,
$
v = f\lambda \\
\therefore v = \dfrac{\omega }{{2\pi }} \times \dfrac{{2\pi }}{k} = \dfrac{\omega }{k} \\
v = \dfrac{{36}}{{0.018}} = 2000\,cm/s = 20\,m/s \\
$
Hence, the speed of the given travelling wave is $20\,m/s$
(B). From the equations (1) and (2), we find that
Amplitude of the given wave, $a = 3\,cm$
Frequency of the given wave, $f = \dfrac{\omega }{{2\pi }} = \dfrac{{36}}{2} \times 3.14 = 573\,Hz$
(C). Again comparing equations (1) and (2), the initial phase angle, $\phi = \dfrac{\pi }{4}$
(D). The least distance between two successive crests in the wave is equal to the wavelength of the wave.
Wavelength, $\lambda = \dfrac{{2\pi }}{k} = \dfrac{{2 \times 3.14}}{{0.018}} = 348.89\,cm = 3.49\,m$
Note: Here we have to remember only the displacement equations otherwise the simplification of this question is not possible. Also since the wave has both frequency and wavelength, we can assume the wave as a travelling wave and not a stationary one.
Then we have to find the frequency of the given wave by using the formula between frequency and angular velocity.
Also we can get the phase angle by comparing the displacement equation and the given equation.
Lastly, we can find the wavelength by determining the distance between the successive crests.
Complete step by step solution:
(A). The displacement equation of wave travelling from left to right is given by:
$y(x,t) = a\sin (\omega t + kx + \phi )$
...... (1)
The given equation in the question is:
$y(x,t) = 3.0\sin (36t + 0.018x + \dfrac{\pi }{4})$
...... (2)
When we compare the two equations, we find that equation (2) also represents the displacement equation of a wave. So, it is a travelling wave.
From equations (1) and (2) we get,
$\omega = 36\,rad/s$, $k = 0.018\,{m^{ - 1}}$
We know that,
Speed, $v = \dfrac{\omega }{{2\pi }}$ and wavelength, $\lambda = \dfrac{{2\pi }}{k}$
Also,
$
v = f\lambda \\
\therefore v = \dfrac{\omega }{{2\pi }} \times \dfrac{{2\pi }}{k} = \dfrac{\omega }{k} \\
v = \dfrac{{36}}{{0.018}} = 2000\,cm/s = 20\,m/s \\
$
Hence, the speed of the given travelling wave is $20\,m/s$
(B). From the equations (1) and (2), we find that
Amplitude of the given wave, $a = 3\,cm$
Frequency of the given wave, $f = \dfrac{\omega }{{2\pi }} = \dfrac{{36}}{2} \times 3.14 = 573\,Hz$
(C). Again comparing equations (1) and (2), the initial phase angle, $\phi = \dfrac{\pi }{4}$
(D). The least distance between two successive crests in the wave is equal to the wavelength of the wave.
Wavelength, $\lambda = \dfrac{{2\pi }}{k} = \dfrac{{2 \times 3.14}}{{0.018}} = 348.89\,cm = 3.49\,m$
Note: Here we have to remember only the displacement equations otherwise the simplification of this question is not possible. Also since the wave has both frequency and wavelength, we can assume the wave as a travelling wave and not a stationary one.
Recently Updated Pages
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x1x2xn be in an AP of x1 + x4 + x9 + x11 + x20-class-11-maths-CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)