Question

A thin taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation

y (x, t) = (5.60 cm) \sin [(0.0340 rad {cm}^{- 1}) x] \sin [(50.0 rad s^{- 1}) t]

, where the origin is at the left end of the string, the

x

-axis is along the string and the

y

-axis is perpendicular to the string.
(a) Draw a sketch that shows the standing wave pattern.
(b) Find the amplitude of the two travelling waves that make up this standing wave.
(c) What is the length of the string?
(d) Find the wavelength, frequency, period and speed of the travelling wave.
(e) Find the maximum transverse speed of a point on the string.
(f) What would be the equation

y (x, t)

for this string if it were vibrating in its eighth harmonic?

Accepted Answer

Given that, a string is tied at both ends, and it if vibrating with its third harmonic wave equation is given as,
$y (x, t) = (5.6 cm) \sin [(0.034 rad {cm}^{- 1}) x] \sin [(50 rad s^{- 1}) t]$
Compare the given equation with, $y = A_{s} \sin k_{3} x \sin ω_{3} t$ .
Amplitude of standing wave is, $A_{s} = 2 A = 5.6 cm$ .
Propagation constant at third harmonic is, $k_{3} = 0.034 rad {cm}^{- 1}$ .
Angular frequency of third harmonic is $ω_{3} = 50 rad s^{- 1}$ .
(a) Given that, the string is vibrating with its third harmonic, i.e, the string is vibrating with three loops.

(b) The amplitude of standing wave,
$2 A = 5.6 cm \Rightarrow A = \frac{5.6}{2} = 2.8 cm$ , here $A$ is the amplitude of each wave.

(c) Wavelength of third harmonic is, $λ_{3} = \frac{2 L}{3} \Rightarrow L = \frac{3 λ_{3}}{2}$

$L = \frac{3}{2} (\frac{2 π}{k_{3}})$ , since, $(k_{3} = \frac{2 π}{λ_{3}} \Rightarrow λ_{3} = \frac{2 π}{k_{3}})$

$L = \frac{3}{2} (\frac{2 (3.14)}{0.034}) ≅ 277 cm = 2.77 m$
Hence, the length of the given wire is, $L = 2.77 m$

(d) Wavelength of third harmonic is, $λ_{3} = \frac{2 L}{3} = \frac{2 (277)}{3} = 184.67 cm$

Speed of travelling wave is, $v = \frac{coefficient of t}{coefficient of x}$
$x and t$ are the displacement of the wave and time taken.
$v = \frac{ω_{3}}{k_{3}} = \frac{50}{0.034} = 1470.59 ≅ 1471 cm s^{- 1}$

Frequency of third harmonic of the wave is, $n_{3} = \frac{ω_{3}}{2 π} = \frac{50}{2 (3.14)} = 7.96 Hz$

Time period third harmonic of the wave is, $T_{3} = \frac{1}{n_{3}} = \frac{1}{7.96} ≅ 0.126 s$

(e) Maximum transverse speed of any particle in third harmonic of the wave is, $V_{\max} = ω_{3} (2 A) = 50 (5.6) = 280 cm s^{- 1}$

(f) If the string vibrating with $8 th$ harmonic, its amplitude remains same, angular frequency and propagation constant changes as, $ω^{'} = 2 π n_{8} = 2 π (8 n_{o}) \times \frac{3}{3} = \frac{8}{3} (2 π (3 n_{o}))$
$ω^{'} = \frac{8}{3} ω_{3} = \frac{8}{3} (50) ≅ 133 rad s^{- 1}$

$λ_{8} = \frac{2 L}{8} \times \frac{3}{3} = \frac{3}{8} (\frac{2 L}{3}) = \frac{3}{8} λ_{3} = \frac{3}{8} (184.67) = 69.25 cm$
$k^{'} = \frac{2 π}{λ_{8}} = \frac{2 (3.14)}{69.25} ≅ 0.091 rad {cm}^{- 1}$
Equation of wave corresponds to $8 th$ harmonic is,
$y^{'} (x, t) = (5.6 cm) \sin (k^{'} x) \sin (ω^{'} t) y^{'} (x, t) = (5.6 cm) \sin ((0.091 rad {cm}^{- 1}) x) \sin ((133 rad s^{- 1}) t)$

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