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PtichkaEL [24]
3 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
Physics
1 answer:
alukav5142 [94]3 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.

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Answer:

The frequency f = 521.59 Hz

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Explanation:

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Diameter of the tank = 44 cm

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Diameter of the spigot = 3.0 mm

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From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

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P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

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g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

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v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

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v_s (velocity of sound) = 343 m/s

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h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

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∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

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