A sinusoidal wave is traveling on a string with speed 74.5 cm/s. The displacemen

Question

A sinusoidal wave is traveling on a string with speed 74.5 cm/s. The displacement of the particles of the string at x = 17 cm is found to vary with time according to the equation y = (3.2 cm) sin[1.3-(5.7 s-1)t]. The linear density of the string is 2.1 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form y(x,t)-Ym sin(kx - ot), what are (c) ym» (d) k, and (e) a, and (f) the correct choice of sign in front of a? (g) what is the tension in the string? The inear bersty of the string i 2.i g/em. What are (a) the requency and (b) the wavelength of the wave? Il the wave equation is of the fom (a) Number (b) Number (c) Number (d) Number (e) Number Units Units Units Units Units (9) Number Units

Explanation / Answer

(a)

Angular velocity from given equation is w = 5.7 s^-1

Then the frequency of the wave is f = w / 2*pi = 5.7 / ( 2*3.14 ) = 0.908 Hz = 0.91 Hz

(b)

The relation between frequency and velocity is,

v = f * lambada

Then the wavelength is,

lambada = v / f

lambada = 74.5 cm/s / 0.908 Hz

lambada = 0.821 m (or) 82.1 cm

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(c)

Compare the both equations, then you will get

ym = 3.2 cm

(d)

Compare the both equations, then you get

k*x = 1.3

k = 1.3 / x

k = 1.3 / 0.17

k = 7.65 N/m

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(e)

Angular velocity from given equation is w = 5.7 s^-1

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(f)

sign of w is positive (+)

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(g)

velocity of the wave in a string is,

v = sqrt (T / mu)

v^2 = T / mu

T = mu * v^2

T = 2.1 g / cm * (74.5 cm/s) ^2

T = 0.21 *0.745^2

T = 0.117 N

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