Question

A harmonic wave with a frequency of 80HZ and an amplitude of 0.025m travels along a string to the right with a speed of 12m/s.
a) write a suitable wave function of 12m/s.
b) Find the maximum speed of a point on the string.
C) Find the maximum acceleration of a point on the string.

Answer 1

Answer:

A harmonic wave with a frequency of 80Hz and an amplitude of 0.025m travels along a string to the right with a speed of 12m/s. a) Write a suitable wave function for this wave. b) Find the maximum speed of a point on the string. c) Find the maximum acceleration of a point on the string. Two point masses m1 and m2 are separated by a massless rod of length L. a) Write an expression for the moment of inertia about an axis perpendicular to the rod and passing through it at a distance x from mass m1. b) Calculate dI/dx and show that I is at a minimum when the axis passes through the center of mass of the system.

Explanation:

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Answer 2

Answer:

The wave equation  is

$y(x,t)=0.025sin2\pi \bigg(80t-\dfrac{x}{0.15} \bigg)$

Maximum speed and acceleration are 12.56m/s and $6316.42m/s^{2}$ respectively.

Explanation:

Given the frequency of a harmonic wave, n = 80HZ

Amplitude, A = 0.025m

Speed of the wave, v = 12m/s.

a) Given the waves travels to the right, the wave function defined by

$y=f\bigg(t-\dfrac{x}{\lambda} \bigg)$

and thus the sinusoidal harmonic wave travelling in the positive direction is given by

$y(x,t)=Asin2\pi \bigg(nt-\dfrac{x}{\lambda} \bigg)$

where $\omega=2\pi n$ is the angular frequency, t is the time, x is the displacement and  $k=2\pi/\lambda$ is the wave number.

From the relation of the velocity of a wave, $v=n\lambda$'

$\lambda=\dfrac{v}{n} =\dfrac{12}{80} =0.15m$

Hence the wave equation  is given by

$y(x,t)=0.025sin2\pi \bigg(80t-\dfrac{x}{0.15} \bigg)$

b) Maximum speed of a point on the string is at the maximum amplitude.

i.e., $v=A \omega$

Angular frequency, $\omega=2\pi n=2\times 3.14\times 80=502.65rad/s$

Therefore, the maximum speed is

$v= 0.025 \times502.65=12.56m/s$

c) Maximum acceleration of a point on the string is at the maximum amplitude.

i.e., $a=A \omega^2$

Therefore, the maximum acceleration is

$a= 0.025 \times(502.65)^{2} =6316.42m/s^{2}$

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