(a) A string with both ends held fixed is vibrating in its third harmonic. The w

Question

(a) A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 56.0 m/s and a frequency of 80.0 Hz. The amplitude of the standing wave at an antinode is 0.700 cm. Calculate the amplitude at points on the string at the following distances.

(i) 35 cm from the left end of the string

(ii) 17.5 cm from the left end of the string

(iii) 8.75 cm from the left end of the string

(b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?

(i) 35 cm from the left end of the string

(ii) 17.5 cm from the left end of the string

(iii) 8.75 cm from the left end of the string

(c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points in part (a).

(i) 35 cm from the left end of the string

(ii) 17.5 cm from the left end of the string

(iii) 8.75 cm from the left end of the string

Explanation / Answer

a)

Equation of a standing wave is:

y = y0 sin(kx) cos(t)

= 2f = 2*3.14*80.0 = 502.4 rad/s

= c/f = 56/80 = 0.700 m

k = 2/ = 2/0.700

y0 = 0.700 cm

>>> y = y0 sin(kx) cos(t)

>>> y = 0.700 sin((2/0.700) * x) cos(502.4 * t)

At poit x, the amplitude is y0 sin(kx)

>>>> Amplitude = 0.700 sin((2/0.700)*x)

(i) Amplitude at x=35 cm from the left end of the string

Amplitude = 0.700*sin((2/0.700)*0.35) = 0 cm

(ii) Amplitude at x=17.5 cm from the left end of the string

Amplitude = 0.700*sin((2/0.700)*0.175) = 0.700 cm

(iii) Amplitude at x=8.75 cm from the left end of the string

Amplitude = 0.700*sin((2/0.700)*0.0875) = 0.495 cm

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b)

Period of the all points is T:

T = 1/f = 1/80.0 = 0.0125 s

time does it take the string to go from its largest upward displacement to its largest downward displacement = T/2 = 0.00625 s

(i) 35 cm from the left end of the string

0.00625 s

(ii) 17.5 cm from the left end of the string

0.00625 s

(iii) 8.75 cm from the left end of the string

0.00625 s

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c)

v = dy/dt = -y0 sin(kx) sin(t)

>>> At poin x, the maximum transverse velocity is y0 sin(kx)

a = dv/dt = -2y0 sin(kx) cos(t)

>>> At poin x, the maximum transverse acceleration is 2y0 sin(kx)

>>> vmax = y0 sin((2/0.700)*x)

>>> amax = 2y0 sin((2/0.700)*x)

(i) 35 cm from the left end of the string

>>> vmax = 502.4*0.700*sin((2/0.700)*0.35) = 0

>>> amax = 502.42*0.700*sin((2/0.700)*0.35) = 0

(ii) 17.5 cm from the left end of the string

>>> vmax = 502.4*0.700*sin((2/0.700)*0.175) = 352 m/s

>>> amax = 502.42*0.700*sin((2/0.700)*0.175) = 1.77 * 10^5 m/s2

(iii) 8.75 cm from the left end of the string

>>> vmax = 502.4*0.700*sin((2/0.700)*0.0875) = 249 m/s

>>> amax = 502.42*0.700*sin((2/0.700)*0.0875) = 1.25 *10^5 m/s2

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