A sinusoidal transverse wave of wavelength 20 cm travels along a string in the positive direction of an $$x$$ axis. The displacement $$y$$ of the string particle at $$ x = 0$$ is given in Fig. 16-34 as a function of time $$t$$. The scale of the vertical axis is set by $$y_s = 4.0 \space cm$$. The wave equation is to be in the form $$y\left (x, t \right ) = y_m \sin \left (kx \pm \omega t + \phi \right )$$. (a) At $$t = 0$$, is a plot of $$y$$ versus $$x$$ in the shape of a positive sine function or a negative sine function? What are (b) $$y_m$$ (c) $$k$$ (d) $$\omega$$ (e) $$\phi$$ (f) the sign in front of $$\omega$$ and (g) the speed of the wave? (h) What is the transverse velocity of the particle at $$x = 0$$ when $$t = 5.0 \space s$$?
A general expression for a sinusoidal wave traveling along the $$+x$$ direction is
$$y\left (x, t \right ) = y_m \sin \left (kx - \omega t + \phi \right )$$.
(a) The figure shows that at $$ x = 0, y\left (0, t \right ) = y_m \sin \left (- \omega t + \phi \right )$$ is positive sine function, that is $$y\left (0, t \right ) = + y_m \sin \omega t $$. Therefore, the phase constant must be $$\phi = \pi$$. At $$t = 0$$, we then have
$$y\left (x, 0 \right ) = y_m \sin \left (kx + \phi \right ) = - y_m \sin kx$$
which is a negative sine function. A plot of $$y\left (x, 0 \right )$$ is depicted on the right.
(b) From the figure we see that the amplitude is $$y_m = 4.0 \space cm$$
(c) The angular wave number is given by $$k = 2\pi/\lambda = \pi/10 = 0.31 \space rad/cm$$.
(d) The angular frequency is $$\omega = 2\pi/T = \pi/5 = 0.63 \space rad/s$$
(e) As found in part (a), the phase is $$\phi = \pi$$.
(f) The sign is minus since the wave is traveling in the $$+x$$ direction.
(g) Since the frequency is $$f = 1/T = 0.10 \space s$$, the speed of the wave is $$v = f\lambda = 2.0 \space cm/s$$
(h) From the results above, the wave may be expressed as
$$y\left (x, t \right )= 4.0 \sin \left (\dfrac{\pi x} {10} - \dfrac{\pi t} {5}+ \pi \right ) = -4.0 \sin \left (\dfrac{\pi x} {10} - \dfrac{\pi t} {5} \right )$$.
Taking the derivative of y with respect to $$t$$, we find
$$u\left (x, t \right ) = \dfrac{\partial y} {\partial t} = 4.0\left (\dfrac{\pi} {t} \right ) \cos \left (\dfrac{\pi x} {10} - \dfrac{\pi t} {5} \right )$$
which yields $$u\left (0, 5.0 \right ) = -2.5 \space cm/s$$.
A string of length 40 cm and weighing 10 g is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of $$160 N m^{-1}$$ and is stretched by 1.0 cm. If a wave pulse is produced on the string near the wall, how much time will it take to reach the spring ?
The equation of a wave travelling on a string stretched along the X-axis is given by $$y = Ae^{\left(\dfrac{x}{a} + \dfrac{t}{T} \right)^2}$$ Find the wave speed.
The equation of a travelling wave is $$y=60\cos (1800\ t-6x)$$ where $$y$$ is in microns, $$t$$ in seconds and $$x$$ in metres. The ratio of maximum particle velocity to velocity of wave propagation is:
the mean value of its velocity vector projection $$\left<\nu_n \right>$$;
the modulus of the mean velocity vector $$|\left
The linear density of a string is $$1.6 \times 10^{-4} \space kg/m$$. A transverse wave on the string is described by the equation $$y = \left (0.021 \space m \right ) \sin \left [\left (2.0 \space m^{-1} \right ) x + \left (30 \space s^{-1} \right ) t \right ]$$. What are (a) the wave speed and (b) the tension in the string?
A generator at one end of a very long string creates a wave given by $$y=(6.0 \mathrm{cm}) \cos \dfrac{\pi}{2}\left[\left(2.00 \mathrm{m}^{-1}\right) x+\left(8.00 \mathrm{s}^{-1}\right) t\right]$$ and a generator at the other end creates the wave $$y=(6.0 \mathrm{cm}) \cos \dfrac{\pi}{2}\left[\left(2.00 \mathrm{m}^{-1}\right) x-\left(8.00 \mathrm{s}^{-1}\right) t\right]$$ Calculate the speed of each wave.
Use the wave equation to find the speed of a wave given by $$y\left ( x, t \right ) = \left (3.00 \space mm \right ) \sin \left [\left (4.00 \space m^{-1} \right ) x - \left ( 7.00 \space s^{-1} \right )t \right ]$$
Use the wave equation to find the speed of a wave given by $$y\left ( x, t \right ) = \left (2.00 \space mm \right ) \left [\left (20 \space m^{-1} \right ) x - \left (4.00 \space s^{-1} \right )t \right ]^{0.5}$$
A sinusoidal transverse wave traveling in the positive direction of an $$ x $$ axis has an amplitude of $$ 2.0 \mathrm{cm}, $$ a wavelength of $$ 10 \mathrm{cm}, $$ and a frequency of $$ 400 \mathrm{Hz} $$. If the wave equation is of the form $$ y(x, t)=y_{m} \sin (k x \pm \omega t), $$ what is the speed of the wave?