What increases at the speaker with the increase of loudless-intensity?

  • Thread starter Thread starter luckis11
  • Start date Start date
  • Tags Tags
    increase Speaker
AI Thread Summary
Maximum displacement, or amplitude, directly affects loudness, with larger amplitudes resulting in louder sounds. Speed of movement increases when displacement increases within the same time frame. The frequency of vibrations, which corresponds to musical notes, can increase or decrease depending on the system's parameters. Understanding these relationships is essential for grasping the physics of sound. Overall, the discussion emphasizes the interconnectedness of amplitude, speed, and frequency in sound production.
luckis11
Messages
272
Reaction score
2
Surely, its maximum displacement A increases.

But what about its speed? Increases, decreases, remains the same?

And what about the number of the vibrations of its maximum displacement per sec? They increase or decrease or remain the same? This number is the frequency of the music f, or not?

Please post only those who are certain of the answer, or certain of what today's physics claim on this.
 
Last edited:
Physics news on Phys.org
The number of vibrations per second is the musical note.

The loudness is determined by the amplitude (size) of the movement.

So, if you want something to move further in the same amount of time, you have to move it faster.

It ain't Physics - it's just plain common sense.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top