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Dmitry_Shevchenko [17]
3 years ago
14

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) x] sin [(50.0 rad/s) 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 waved 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 waves.
(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?
Physics
1 answer:
barxatty [35]3 years ago
4 0

Answer:

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation y(x,t)=(5.60cm)sin[(0.0340rad/cm)x]sin[(50.0rad/s)t]y(x,t)=(5.60cm)sin[(0.0340rad/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 y-axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling 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 traveling waves. (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?

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Answer:

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Explanation:

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You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

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The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

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