A this, taut string tied at both ends and oscillating in its Hard harmonic has i

Question

A this, taut string tied at both ends and oscillating in its Hard harmonic has its shape described by the equation y(x, t) = (5-60cm)sin[(0.0340 rad/cm)x]sin[(50.0rad/s)t}, where the origin is at the left end of the string, the x-axis is along the string and the yaxis is perpendicular to the string, Draw a sketch that shows the standing wave pattern, Hud the amplitude of the two traveling waves that make up this standing wave, What is the length of the string? Hud the wavelength, frequency, period, and speed of the traveling waves, find the maximum transverse speed of a point on the string, What would be the equation y( x, t) for this string if it were vibrating in its eighth harmonic?

Explanation / Answer

given equation

y(x,t) = (5.60cm) sin [(0.0340 rad/cm)x] sin [(50.0 rad/s)t]

and a standing wave equation in third harmonic will be ,

y = 2A sin kx sin t 3 L

now, we start comparing ,

(a)

y (for standing wave) = 2A sin kx sin t Thus, by comparison to the given equation, 2A = 5.6 cm so A = 2.8 cm

(b) 3 = 2L/3 or L = (3/2) 3

or, f3 = 3(v/2L) = v/3.

But the wavenumber k gives 2/ ,

k = 0.034 rad/cm = 3.4/m so that = 2/k = 2/(3.4/m) = 1.85 m. Thus L = (3/2) 3 = 2.78 m

(c) The wavelength of traveling waves is the same as the standing wave: = 1.85 m

(d) The frequency and period of the traveling waves is the same as the standing wave.

= 50 rad/s or f = /2 = 7.96 Hz. and T = 1/f = 0.1256 sec

(e) The speed of the traveling wave is given by v = f = (2.78 m)(7.96Hz) = 22.13 m/s

(f) Maximum transverse speed of a point on the string can be found by taking dy/dt and maximixing this. dy/dt = (5.60 cm)(50 rad/s) sin [(0.0340rad/cm)x] [cos (50 rad/s)t]. T

the maximum dy/dt must be (5.6 cm)(50 rad/s) = 280 rad/

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.