Question

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

Answer 1

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.