anaylse of travelling wavs of physics, include question and solution

1. Travelling Waves
Yukun Fang

2. • Wave motion is very different from the motion of objects.
Some waves need a medium in which to travel, such as waves
in ocean, but some do not, such as light.
• Wave Speed: The speed at which a disturbance travels
through a medium is called the wave speed, v
• Transverse Waves: The constituents of the medium move
perpendicular to the direction of propagation of the wave in a
transverse wave.
• Longitudinal Waves: In a longitudinal wave, the constituents
of the medium move parallel to the direction of motion of the
wave.

3. • Waves carry energy: The pulse carries this energy as it travels
along the sting. It takes energy to generate a pulse, and the
pulse carries this energy as it propagates through the medium.
A wave can be described as the motion of energy through a
medium.
• The medium does not travel with the wave: The pulse always
pass through its equilibrium position and then returns to the
equilibrium position. There is no net movement of the string
in the direction of motion of the pulse. (eg, a sound wave
does not carry air along with it)
• Equation of a Pulse Moving in One Dimension: The transverse
displacement of an element of the string form its equilibrium
position is denoted by D(x, t); where x is displacement of the
transverse wave, t is time. A function of the form D(x+/-vt) is
called a wave function.

4. • Wave Speed on a String: 𝜇=M/L, where M is the total mass
of a string; L is the length of the string.
• F=ma, a=v2/R, m*(
𝑣2
𝑅
)=Ts(
𝐿
𝑅
), therefore, v=√
𝑇𝑠
𝜇

5. • Question: A string of linear mass density 10g/m and length
10m is hanging vertically from a high ceiling. A mass of 1kg
hangs freely from the lower end of the string. A pulse is
generated at the lower end of the string and travels upward.
What is the speed of the pulse ate the lower end of the string,
at the middle of the string, and at the top of the string?
• Solution: use the equation v(x)=√
𝑀𝑔+𝜇𝑥𝑔
𝜇
because the tension
force in this case is the total weigh hanging below the point x,
then, evaluating the wave speed at x=0, x=5m,and x=10m
• We can get: v(0)=√
𝑀𝑔
𝜇
= √
1𝑘𝑔∗9.81𝑚/𝑠
10∗10−3 = 31.30m/s;
v(5)=√
𝑀𝑔+5𝜇𝑔
𝜇
= √
1∗9.8+5∗10∗10−3∗9.8
10∗10−3 = 32.08m/s
v(10)=√
𝑀𝑔+10𝜇𝑔
𝜇
=√
1∗9.8+10∗10∗10−3∗9.8
10∗10−3 = 32.83m/s

6. Thank you!