Your attempt looks good except for the first one.

  • Thread starter Thread starter vande060
  • Start date Start date
  • Tags Tags
    Antenna Waves
AI Thread Summary
The discussion focuses on calculating the wavelengths of the four longest standing waves on a 1m automobile antenna, which is fixed at one end and free at the other. The correct formula for the wavelengths is λn = (2/n) * L, where L is the length of the antenna. The user initially provides incorrect calculations for the wavelengths, listing them as 2m, 1m, 2/3m, and 0.5m. It is emphasized that one end of the antenna acts as a node and the other as an antinode, suggesting that visualizing the standing waves can aid in understanding the problem. The importance of correctly applying the standing wave principles is highlighted for accurate results.
vande060
Messages
180
Reaction score
0

Homework Statement


Your automobile antenna is 1m in length and is fixed at one end, free at the other. When you drive on an
uneven road, waves can be excited on the antenna. Compute the wavelengths of the four longest wavelength standing
waves that your antenna will allow.

Homework Equations


for a standing wave i know these equations exist

λn =(2/n )*L

also A' = -A

furthermore, I know that the fixed end of the antenna could be considered to have infinite density

The Attempt at a Solution



the four longest wavelengths should be:

2/1 * 1 = 2 m
2/2 * 1 = 1 m
2/3 * 1 = 2/3m
2/4 * 1 = .5m

is this even close?
 
Physics news on Phys.org
vande060 said:

Homework Statement


Your automobile antenna is 1m in length and is fixed at one end, free at the other. When you drive on an
uneven road, waves can be excited on the antenna. Compute the wavelengths of the four longest wavelength standing
waves that your antenna will allow.

If it is fixed at one end and free at the other, one end must be a node, and the other end must be an antinode.

Try to draw what the acceptable standing waves would look like; that will help you to find their wavelengths.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top