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andreyandreev [35.5K]
3 years ago
15

For a given wave IF frequency doubles the wavelength

Physics
1 answer:
motikmotik3 years ago
7 0

Answer:

C is halved

Explanation:

The frequency and the wavelength of a wave are related by the equation:

v=f\lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

From the equation above, we see that for a given wave, if the wave is travelling in the same medium (and so, its speed is not changing), then the frequency and the wavelength are inversely proportional to each other.

Therefore, if the frequency doubles, the wavelength will halve in order to keep the speed constant:

\lambda = \frac{v}{f}\\\lambda' = \frac{v}{f'}=\frac{v}{2f}=\frac{1}{2}(\frac{v}{f})=\frac{\lambda}{2}

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If the primary source of atmospheric thermal energy on Earth is energy from the Sun, what conclusion can be made from the diagra
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Which of the following statement(s) about energy and phase is/are correct? Select all that apply. Choose one or more: A. While o
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Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in
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Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

v^{2} = \frac{dr}{dt}^{2}  +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2} - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

\frac{r}{z}=tan(45)

or:

z = r cot(45) - (2)

and:

\frac{dz}{dt} = \frac{dr}{dt} cot(45)

replacing (2) in (1) we obtain:

v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}  - (3)

Now the kinetic energy is given as:

T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}) - (4)

And the potential energy is given by:

V = -mgz = -mgr cot(45)

So the Langrangian is given by:

L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)

And the equations of motion are:

For θ

\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c

For r

\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

mr\frac{d\theta}{dt}=c

Which is the magnitude of the angular momentum

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