Ask question

Ask question

All categories

4 years ago

11

A transverse wave on a string is described by the following wave function.y = (0.090 m) sin (px/11 + 4pt)(a) Determine the trans

verse speed and acceleration of an element of the string at t = 0.160 s for the point on the string located at x = 1.40 m.Your response differs from the correct answer by more than 10%. Double check your calculations. m/sm/s2(b) What are the wavelength, period, and speed of propagation of this wave?msm/s

1 answer:

alukav5142 [94]4 years ago

7 0

Explanation:

(a) It is known that equation for transverse wave is given as follows.

y = $(0.09 m)sin(\pi \frac{x}{11} + 4 \pi t)$

Now, we will compare above equation with the standard form of transeverse wave equation,

y = $A sin(kx + \omega t)$

where, A is the amplitude = 0.09 m

k is the wave vector = $\frac{\pi}{11}$

$\omega$ is the angular frequency = $4\pi$

x is displacement = 1.40 m

t is the time = 0.16 s

Now, we will differentiate the equation with respect to t as follows.

The speed of the wave will be:

v(t) = $\frac{dy}{dt}$

v(t) = $A \omega cos(kx + \omega t)$

v(t) = $(0.09 m)(4\pi) cos(\frac{\pi \times 1.4}{11} + 4 \pi \times 0.16)$

v(t) = -0.84 m/s

The acceleration of the particle in the location is

a(t) = $\frac{dv}{dt}$

a(t) = $-A \omega 2sin(kx + \omega t)$

a(t) = $-(0.09 m)(4 \pi)2 sin(\frac{\pi \times 1.4}{11} + 4\pi \times 0.16)$

a(t) = -9.49 $m/s^{2}$

Hence, the value of transverse wave is 0.84 m/s and the value of acceleration is 9.49 $m/s^{2}$ .

(b) Wavelength of the wave is given as follows.

$\lambda = \frac{2\pi}{k}$

$\lambda = (frac{2\pi}{\frac{\pi}{11})
$

$\lambda$ = 22 m

The period of the wave is

T = $\frac{2 \pi}{\omega}$

T = $\frac{2 \pi}{4 \pi}$

= 0.5 sec

Now, we will calculate the speed of propagation of wave as follows.

v = $\frac{\lambda}{T}$

= $\frac{22 m}{0.5 s}$

= 44 m/s

therefore, we can conclude that wavelength is 22 m, period is 0.5 sec, and speed of propagation of wave is 44 m/s.

You might be interested in

Yolanda is performing a neutralization reaction using hydrochloric acid and sodium hydroxide. Since both are clear liquids, what

Bad White [126]

I should think that Yolanda should use litmus paper throughout the reaction as its color will tell when the mixture is neutral and therefore then, when she can stop adding either the hydrochloric acid or the sodium hydroxide or if one or the other needs to be added to make the solution approach neutral. Blue litmus paper stays blue in a neutral solution, but will turn red in an acidic solution.

5 0

4 years ago

To water the yard you use a hose with a diameter of 3.2 cm. Water flows from the hose with a speed of 1.1 m/s. If you partially

cupoosta [38]

Answer:

The speed of water flow inside the pipe at point - 2 = 34.67 m / sec

Explanation:

Given data

Diameter at point - 1 = 3.2 cm

Velocity at point - 1 = 1.1 m / sec = 110 cm / sec

Diameter at point - 2 = 0.57 cm

Velocity at point - 2 = ??

We know that from the continuity equation the rate of flow is constant inside a pipe between two points.

Thus

⇒ $A_{1}$ × $V_{1}$ = $A_{2}$ × $V_{2}$

⇒ $\frac{\pi }{4}$ × $d_{1} ^{2}$ × $V_{1}$ =

⇒ $d_{1} ^{2}$ × $V_{1}$ = $d_{2} ^{2}$ × $V_{2}$

⇒ $(3.2)^{2}$ × 110 = $(0.57)^{2}$ × $V_{2}$

⇒ $V_{2}$ = 3467 cm / sec

⇒ $V_{2}$ = 34.67 m / sec

Thus the speed of water flow inside the pipe at point - 2 = 34.67 m / sec

3 0

3 years ago

Help asapppppppppppppppppp

Lana71 [14]

Answer:

I’m pretty sure the answer is distance

Explanation:

Hope this helps! Sorry if it’s wrong.

5 0

3 years ago

There are 3600 in a full circle. a. How many arcminutes are in a full cycle? b. How many arcseconds are in a full cycle?

liberstina [14]

Answer:

21600

1296000

Explanation:

Full cycle consists of

$360^\{circ}$

1 hour = 3600 seconds

1 minute = 60 seconds

Converting to arcminutes

$360\times 60=21600\ arcseconds$

The arcminutes in a full cycle is 21600

Converting to arcseconds

$360\times 60\times 60=1296000\ arcseconds$

The arseconds in full cycle is 1296000

4 0

3 years ago

The Earth is 1.5 × 1011 m from the Sun and takes a year to make one complete orbit. It rotates on its own axis once per day. It

katrin2010 [14]

Answer:

Part a)

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

$KE_{tot} = 2.68 \times 10^{33} J$

Explanation:

Time period of Earth about Sun is 1 Year

so it is

$T = 1 year = 3.15 \times 10^7 s$

now we know that angular speed of the Earth about Sun is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{3.15 \times 10^7}$

now speed of center of Earth is given as

$v_{cm} = r\omega$

$r = 1.5 \times 10^{11} m$

$v_{cm} = (1.5 \times 10^{11})(\frac{2\pi}{3.15 \times 10^7})$

$v_{cm} = 2.98 \times 10^4 m/s$

Part b)

now transnational kinetic energy of center of Earth is given as

$K_{trans} = \frac{1}{2}mv^2$

$K_{trans} = \frac{1}{2}(6 \times 10^{24})(2.98 \times 10^4)^2$

$K_{trans} = 2.68 \times 10^{33} J$

Part c)

Angular speed of Earth about its own axis is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{24 \times 3600}$

$\omega = 7.27 \times 10^{-5} rad/s$

Part d)

Now moment of inertia of Earth about its own axis

$I = \frac{2}{5}mR^2$

$I = \frac{2}{5}(6 \times 10^{24})(6.4 \times 10^6)^2$

$I = 9.83 \times 10^{37} kg m^2$

now rotational energy is given as

$KE_{rot} = \frac{1}{2}I\omega^2$

$KE_{rot} = \frac{1}{2}(9.83 \times 10^{37})(7.27 \times 10^{-5})^2$

$KE_{rot} = 2.6 \times 10^{29} J$

Part e)

Now total kinetic energy is given as

$KE_{tot} = KE_{trans} + KE_{rot}$

$KE_{tot} = 2.68 \times 10^{33} + 2.6 \times 10^{29}$

$KE_{tot} = 2.68 \times 10^{33} J$

6 0

3 years ago