Answer:
a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).
b. third harmonic
c. to calculate frequency , we compare with general wave equation
y(x,t)=Acos(kx+ωt)
from ωt=742t
ω=742
ω=2*pi*f
742/2*pi
f=118.09Hz
Explanation:
A fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)=2.30mmcos[(6.98rad/m)x+(742rad/s)t]. Being more practical-minded, you measure the rope to have a length of 1.35 m and a mass of 3.38 grams. Assume that the ends of the rope are held fixed and that there is both this traveling wave and the reflected wave traveling in the opposite direction.
A) What is the wavefunction y(x,t) for the standing wave that is produced?
B) In which harmonic is the standing wave oscillating?
C) What is the frequency of the fundamental oscillation?
a. y(x,t)= 2.05 mm cos[( 6.98 rad/m)x + (744 rad/s).
b. lambda=2L/n
when comparing the wave equation with the general wave equation , we get the wavelength to be
2*pi*x/lambda=6.98x
lambda=0.9m
we use the equation
lambda=2L/n
n=number of harmonics
L=length of string
0.9=2(1.35)/n
n=2.7/0.9
n=3
third harmonic
c. to calculate frequency , we compare with general wave equation
y(x,t)=Acos(kx+ωt)
from ωt=742t
ω=742
ω=2*pi*f
742/2*pi
f=118.09Hz
Answer:
Explanation:
graph would be a straight line from (0, 0) to (400, 8)
Plot points are
PE = mgh
50(0) = 0 J
50(2) = 100 J
50(4) = 200 J
50(6) = 300 J
50(8) = 400 J
The resultant static friction force is equal to 20 N to the left.
Why?
I'm assuming that you forgot to write the question of the exercise, so, I will try to complete it:
"A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-n force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?"
So, if we are going to calculate the resulting static friction force, it means that there is no movement, we must remember that the friction coefficient will give us the maximum force before the crate starts to move.
We can calculate the static friction force by using the following formula:

Since the crate is not moving (static), the static friction force acting on the crate will be equal to the applied force.
Calculating we have:


Hence, the static friction force is equal to 20 N to the left (since the applied force is acting to the right)
So,
Since the static friction force is equal to the applied force, the crate does not start to move.
Have a nice day!
Answer:
70 cm
Explanation:
0.5 kg at 20 cm
0.3 kg at 60 cm
x = Distance of the third 0.6 kg mass
Meter stick hanging at 50 cm
Torque about the support point is given by (torque is conserved)

The position of the third mass of 0.6 kg is at 20+50 = 70 cm
Newton's third law says:
"<span>For every action, there is an equal and opposite reaction. ".
So, the force that Tom does on the sister is equal to force the sister applies on Tom:
</span>

<span>where the label "t" means "on Tom", while the label "s" means "on the sister".
From Newton's second law, we also know
</span>

where m is the mass and a the acceleration. <span>so we can rewrite the first equation as
</span>

<span>And find Tom's acceleration:
</span>

<span>
</span>