Ask question

Ask question

All categories

denis-greek [22]

1 year ago

7

A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at

38.0 cm/s. Find its (a) angular frequency and (b) wave number. (c) Write a mathematical expression describing the displacement y of this wave (in centimeters) as a function of position and time. Assume the maximum displacement occurs when t = 0.

1 answer:

storchak [24]1 year ago

4 0

Answer

given,

wavelength = λ = 18.7 cm

= 0.187 m

amplitude , A = 2.34 cm

v = 0.38 m/s

A) angular frequency = ?

$f = \dfrac{v}{\lambda}$

$f = \dfrac{0.38}{0.187}$

$f =2.03\ Hz$

angular frequency ,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) the wave number ,

$K = \dfrac{2\pi}{\lambda}$

$K= \dfrac{2\pi}{0.187}$

$K =33.59\ m^{-1}$

C)

as the wave is propagating in -x direction, the sign is positive between x and t

y ( x ,t) = A sin(k x - ω t)

y ( x ,t) = 2.34 x sin(33.59 x - 12.75 t)

You might be interested in

A motion sensor is used to create the graph of a student’s horizontal velocity as a function of time as the student moves toward

Marrrta [24]

Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

Area of trapezoidal are=

= $\frac{1}{2} *(1.5+0.75)+\frac{1}{2} (1+0.75)(-0.5)$

= $\frac{1}{2} *(2.25-1.75*0.5)$

=0.6875 m

As the area is positive therefore displacement from xo is positive

ax=(change in velocity)/(Time)

ax= $\frac{-0.5-0}{3} =-\frac{1}{6} ms2$

3 0

2 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 years ago

Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a

Phantasy [73]

The question is incomplete. It comes with a set of answer choices.

These are the answer choices:

Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.

Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.

Both Alex and his friend observe it as 10 m/s.

Both Alex and his friend observe it as less than 10 m/s.

Answer: Both Alex and his friend observe it as 10 m/s.

Justification:

1) The speed is relative to the frame of reference.

2) It is said that the both Alex and his friend are standing still.

3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).

Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0

2 years ago

Read 2 more answers

Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0

2 years ago

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

5 0

2 years ago