To solve the problem it is necessary to apply the concepts related to wavelength.
By definition we know that the wavelength is expressed as
[tex]\lambda = \frac{v}{f}[/tex]
Where,
v= velocity
f = frequenciy
On the other hand we have that the distance when there is vibration can be calculated as
[tex]\Delta x = n\lambda[/tex]
Where n is an integer for standing wave pattern, or harmonic, number.
Part A ) The distance between the two objects is basically
[tex]\Delta x = 6-x[/tex]
[tex]\Delta x = 6.2x[/tex]
The velocity is
[tex]v=f\lambda[/tex]
[tex]\lambda = \frac{v}{f}[/tex]
[tex]\lambda = \frac{4m/s}{2Hz}[/tex]
[tex]\lambda =2m[/tex]
On the one hand the the distance for the no vibration is
[tex]\Delta x = (n+\frac{1}{2})\lambda[/tex]
[tex]\Delta x = (0+\frac{1}{2})\lambda[/tex]
[tex]\Delta x = \frac{\lambda}{2}[/tex]
Replacing the value for x,
[tex]6-2x =\frac{2m}{2}[/tex]
[tex]6-2x=1[/tex]
[tex]x=2.5m[/tex]
On the other hand the minimum distance is:
[tex]\Delta x = (n+\frac{1}{2})\lambda[/tex]
[tex]6-2x=\frac{n+\frac{1}{2}}\lambda[/tex]
Re-arrange to x
[tex]x=2.5-n[/tex]
For positive x, the value of n is [tex]n_{max}=2[/tex]
[tex]x = 2.5-2[/tex]
[tex]x=0.5m[/tex]
Therefore the required distance is 0.5m
PART B) The maximum vibration, the distance is
[tex]\Delta x= n\lambda[/tex]
[tex]6-2x = n(2)[/tex]
For the minimum distance, n is maximum, therefore
[tex]6-2x=2*2[/tex]
[tex]x=1m[/tex]
Therefore the required minimum distance is 1m.
