Answer to Question #285631 in Mechanics | Relativity for john

Question

The wave function associated with standing waves established in a 4.00 m long string is

y = 0.00500 sin(πx)cos(150πt) where x and y are in meters and t is in seconds.

(a) How many half-wavelengths does this pattern exhibit?

(b) What is the fundamental frequency of vibration of the string (in Hz)?

c) If the original frequency is held constant and the tension in the string is increased by a factor of 4, how many half wavelengths are present in the new pattern?

Answer 1

The general wave function of a standing wave in a wire is:

$y = 2A sin(kx)cos(\omega t)$

comparing this general function with the given function we get:

$k=\pi=2\pi/\lambda \implies \lambda =2$ m

$\omega=150\pi=2\pi f\implies f=75$ Hz

a)

The distance between adjacent nodes:

$d=\lambda/2$

number of half-wavelengths:

$N=L/d=2L/\lambda=2\cdot4/2=4$

b)

speed of the wave:

$v=f\lambda=75\cdot2=150$ m/s

The standing wave that has the fundamental frequency fits one-half of a wavelength in

the entire length of the string, so its wavelength is:

$\lambda=2L=8$ m

This wave travels the string with the same speed as the wave that fits 4 loops in the length

of the string, so fundamental frequency is then:

$f=v/\lambda=150/8=18.75$ Hz

c)

Let the initial tension be T₁ , the velocity $v_1=150$ m/s, the new tension $T_2=4T_1$ and

the new velocity v₂ , then:

$v_2=\sqrt{T_2/\mu}=\sqrt{4T_1/\mu}=2\sqrt{T_1/\mu}=2v_1=300$ m/s

The new wavelength is:

$\lambda_2=v_2/f=300/75=4$ m

number of half-wavelengths:

$N=L/(\lambda_2/2)=4/2=2$