Question

A sinusoidal transverse wave is travelling along a string in the negative direction of an x axis. The figure shows a plot of the displacement as a function of position at time t = 0; the y intercept is 4.0 cm. The string tension is 3.3 N, and its linear density is 44 g/m. Find the (a) amplitude, (b) wavelength, (c) wave speed, and (d) period of the wave, (e) Find the maximum transverse speed of a particle in the string. If the wave is of the form y(x, t) = y_m sin(kx - omega t + phi), what are (f) k, (g) omega, and (h) phi, and (i) the correct choice of sign in front of omega?

Answer 1

a). From the plot given we can see that maximum amplitude is,

   (as 1 unit on y axis is equal to 1cm)

b). Wavelength is defined as the distance between successive crests or successive troughs.

Then from the plot, we get wavelength as, $\lambda= 4(10cm)= 40cm$

(as 1 unit on x axis is equal to 10cm)

c). We know that wave speed of strings, $v= \sqrt{\frac{T}{\mu}}$

where T= Tension in the string= 3.3 N

$\mu$ = linear density= 44g/m= 0.044 kg/m

then wave speed will be $v= \sqrt{\frac{3.3}{0.044}}= \sqrt{68.181}= 8.257 m/s$

d). We know that if "f" is the frequency of the wave then we'll have

$v= f\lambda$

and Time Period, $T= \frac{1}{f}$

then we'll have $T= \frac{\lambda}{v}$

using all calculated values in above, $T= \frac{0.40}{8.257}= 0.0484 s$

e). Maximum Transverse Speed is given by, $v_{max}= \omega \times A= ( \frac{2 \pi }{T})A$

   $v_{max}= ( \frac{2 \times 3.14 }{0.0484})(0.05)= 6.487 m/s$

f). If the given equation depicts the given wave and plot, then wavenumber(k) is given by

   $k= \frac{2 \pi }{\lambda}= \frac{2 \times 3.14 }{0.4}= 15.7 m^{-1}$

g). And the angular speed is given by $\omega= \frac{2 \pi }{T}= \frac{2 \times 3.14 }{0.0484}= 129.75s^{-1}$

h). Using above values we can write the equation as

$Y(x,t)= y_msin(kx-\omega t+\phi )$

Now since wave travels in negative x direction, hence

   $Y(x,t)= Asin(-15.7x-129.75t+\phi )$

   $Y(x,t)= Asin(15.7x+129.75t+\phi )$

Then using the value of amplitude and value of displacement at time t=0s,

   $0.04= 0.05sin(15.7(0)+129.75(0)+\phi )$

   $0.8= sin(\phi )$

   $\phi= sin^{-1}(0.8)= 53.130^{\circ}$

i). The sign will become positive in front of $\omega$ as shown above.