Select the correct answer.

As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horiz

Eddi Din [679]

a)E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

The total energy of the system at any point in the motion is equal to the sum of the elastic potential energy of the spring, U, and of the kinetic energy of the mass, K:

E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

b) A= $\sqrt{\frac{2E}{k}}$

The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude is 'A'

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency of the motion

t is the time

$\phi$ is the phase (we can take $\phi$ =0 )

The amplitude of the motion occurs when the displacement of the motion is maximum: x=A. Regarding energy, the mass-spring system is at its maximum displacement (x=A) when all the mechanical energy of the framework is elastic potential energy, so when the kinetic energy is zero:

K= $\frac{1}{2}mv^2$ =0

E= $\frac{1}{2}kA^2\\$ -->(1)

A= $\sqrt{\frac{2E}{k}}$

c) $v_{max}=\omega A$

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E= $\frac{1}{2}mv_{max}^2$

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2$

and solving for $v_{max}$ we find an expression for the maximum speed:

$v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A$

<h2> $v_{max}=\omega A$ </h2>

4 0

3 years ago

Assume that you are considering the purchase of a 20-year, noncallable bond with an annual coupon rate of 9.5%. The bond has a f

DiKsa [7]

Answer:

$1,105.69

Explanation:

For this question we use the Present value formula that is shown on the attachment. Kindly find it below:

Given that,

Future value = $1,000

Rate of interest = 8.4% ÷ 2 = 4.2%

NPER = 20 years × 2 = 40 years

PMT = $1,000 × 9.5% ÷ 2 = $47.50

The formula is shown below:

= -PV(Rate;NPER;PMT;FV;type)

So, after solving this, the maximum price pay for the bond is $1,105.69

5 0

3 years ago

Read 2 more answers

An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance

Lorico [155]

Answer:

The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.

Explanation:

Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:

$F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}$

and since the electric field E in between parallel plates separated a distance d and under a potential difference $\Delta V$ , is given by:

$E=\frac{\Delta\,V}{d}$

then :

$a=\frac{q\,\Delta V}{m\,d}$

We want to find when the particle reaches velocity zero via kinematics:

$v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a$

We replace this time (t) in the kinematic equation for the particle displacement:

$\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}$

Replacing the values with the information given, converting the distance d into meters (0.01 m), using $\Delta V=100\,V$ , and the electron's kinetic energy:

$\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J$

we get:

$\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters$ Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:

0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]

8 0

4 years ago

A cart starts at x = +6.0 m and travels towards the origin with a constant speed of 2.0 m/s. What is it the exact cart position

Ira Lisetskai [31]

Answer:

At the origin (x' = 0 m)

Explanation:

Note: From the question, when the cart travels towards the origin, the magnitude of its exact position reduces with time.

The formula of speed is given as

S = d/t................. Equation 1

Where S = speed of the cart, d = distance covered by the cart over a certain time. t = time taken to cover the distance.

make d the subject of the equation,

d = St ................. Equation 2

Given: S = 2.0 m/s, t = 3.0 s

Substitute into equation 2

d = 2(3)

d = 6 m.

From the above, the cart covered a distance of 6 m in 3 s.

The exact position of the cart = Initial position-distance covered

x' = x-d............ Equation 3

Where x' = exact position of the cart 3 s later, x = initial position of the cart, d = distance covered by the cart in 3.0 s.

Given: x = +6.0 m, d = 6 m.

Substitute into equation 3

x' = +6-6

x' = 0 m.

Hence the cart will be at 0 m (origin) 3 s later

5 0

3 years ago

What is primary succession?

dangina [55]

Answer:

Primary succession is one of two types of biological and ecological succession of plant life, occurring in an environment in which new substrate devoid of vegetation and other organisms usually lacking soil, such as a lava flow or area left from retreated glacier, is deposited.

3 0

3 years ago