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anastassius [24]

3 years ago

9

A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i

n its third OVERTONE, and the equation for the vertical displacement of any point on the string is y(x,t) = (1.22 cm) sin[(14.4 m-1)x] cos[(166 rad/s)t]. (a) What are the frequency and wavelength of the fundamental mode of this string? (b) How long is the string? (c) How fast do waves travel on this string?

1 answer:

Fofino [41]3 years ago

4 0

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

$y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]$ (1)

n = 4

Now,

We know the standard eqn is given by:

$y = AsinKxcos\omega t$ (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = $14.4 m^{- 1}$

$\omega = 166\ rad/s$

where

A = Amplitude

K = Propagation constant

$\omega$ = angular velocity

Now, to calculate the string's wavelength,

(a) $K = \frac{2\pi}{\lambda}$

where

K = propagation vector

$\lambda = \frac{2\pi}{K}$

$\lambda = \frac{2\pi}{14.4} = 0.436\ m$

(b) The length of the string is given by:

$l = \frac{n\lambda}{2}$

$l = \frac{4\times 0.436}{2} = 0.872\ m$

(c) Now, we first find the frequency of the wave:

$\omega = 2\pi f$

$f = \frac{\omega}{2\pi}$

$f = \frac{2\pi}{166} = 26.42\ Hz$

Now,

Speed of the wave is given by:

$v = f\lambda$

$v = 26.419\times 0.436 = 11.518\ m/s$

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