A girl drops a pebble from a high cliff into a lake far below. She sees the splash of the pebble hitting the water 2.00s later.

Response:

A=0.199

Clarification:

We know that  

Mass of spring=m=450 g=

Where 1 kg=1000 g

Frequency of oscillation=

Energy for oscillation is 0.51 J

To determine the amplitude of oscillations.

Energy for oscillator=

Where =Angular frequency

A=Amplitude

Using the formula

Therefore, the amplitude of oscillation=A=0.199

Answer:

The final size is nearly the same as the initial size because the increase in size is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV =

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass,

Initial size of the wave packet,

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

This proton energy is

Thus, the speed of the proton, v

The time to cover 1 km = 1000 m of distance is calculated as:

T =

T =

According to the dispersion factor;

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

where

= final width

Since it's classified as a transverse wave, the particle on the string moves horizontally as the wave progresses, without actual forward or backward travel. Consequently, the red dot shifts 'A' to the left, returns 'A' to the center, moves 'A' to the right, and goes back 'A' to the center once again. Thus, the red dot collectively travels a distance totaling 4A.