Ask question

Ask question

All categories

3 years ago

12

An object oscillates with an angular frequency of 8.0 rad/s. At t = 0, the object is at x0 = 4 cm with an initial velocity v0 =

-25 cm/s. Find the amplitude and the phase constant for the motion.

1 answer:

KatRina [158]3 years ago

6 0

Answer:

$\phi = 0.66 rad$

$A = 5.06 cm$

Explanation:

We have here a simple harmonic motion, so the equation of the position in this motion is:

$x(t)=Acos(\omega t+\phi)$ (1)

A: Amplitude

ω: angular frequency

φ: phase constant

If we take the derivative of x with respect to t from (1), we can find the velocity equation of this motion:

$v(t)=\frac{dx(t)}{dt}=-A\omega sin(\omega t+\phi)$ (2)

Let's evaluate (1) and (2) in t=0.

$x(0)=Acos(\phi)$ (3)

$v(0)=-A\omega sin(\phi)$ (4)

Dividing 4 by 3 we have:

$\frac{v(0)}{x(0)}=-\omega tan(\phi)$

$\phi = tan^{-1}(\frac{-v(0)}{\omega x(0)})$

$\phi = 0.66 rad$

Now, using (3) we can find the amplitude.

$A = \frac{x(0)}{cos(\phi)} = 5.06 cm$

I hope it helps!

You might be interested in

Show the equation of simple pendulum to be dimensionally consistent

nataly862011 [7]

T is in seconds (s)

<span>2pi is dimensionless </span>

<span>L is in meters (m) </span>

<span>g is in meters per second squared (m/s^2) </span>

<span>so you can write the equation for the period of the simple pendulum in its units... </span>

<span>s=sqrt(m/(m/s^2)) </span>

<span>simplify</span>

<span>s=sqrt(m*s^2*1/m) cancelling the m's </span>

<span>s=sqrt(s^2) </span>

<span>s=s </span>

<span>therefore the dimensions on the left side of the equation are equal to the dimensions on the right side of the equation.</span>

6 0

2 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

3 years ago

A submarine is stranded on the bottom of the ocean with its hatch 21.0 m below the surface. calculate the force (in n) needed to

Tanzania [10]

Answer:

28,400 N

Explanation:

Let's start by calculating the pressure that acts on the upper surface of the hatch. It is given by the sum of the atmospheric pressure and the pressure due to the columb of water, which is given by Stevin's law:

$p_{top} = p_{atm} + \rho g h=1.013\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(21.0 m)=3.071 \cdot 10^5 Pa$

On the lower part of the hatch, there is a pressure equal to

$p_{bot}=p_{atm}=1.013\cdot 10^5 Pa$

So, the net pressure acting on the hatch is

$p=p_{top}-p_{bot}=3.071 \cdot 10^5 Pa - 1.013\cdot 10^5 Pa=2.058 \cdot 10^5 Pa$

which acts from above.

The area of the hatch is given by:

$A=\pi r^2 = \pi (\frac{0.420 m}{2})^2=0.138 m^2$

So, the force needed to open the hatch from the inside is equal to the pressure multiplied by the area of the hatch:

$F=pA=(2.058\cdot 10^5 Pa)(0.138 m^2)=28,400 N$

8 0

3 years ago

Magnetic fields exist

Effectus [21]

Answer:

Magnetic fields exist near a magnet, farther away from a magnet, and within a magnet.

So, the answer is D. All of the above.

Let me know if this helps!

6 0

3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

2 years ago