The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as
Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as
Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:
If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well
Replacing 
for n and 202Hz for 
The frequency of the tightened is 205Hz
For the first one 320
second
1200W
Data
R = 12 Ω ∆V = 120V I =? P =?
Solution:
According to Ohm’s law,
∆V = I R
I = ∆V / R
= 120 / 12
= 10 A
Power P = I ∆V
= 10 x 120
= 1200 W
Third
∆V = 120 V P = 60 W I =? R =?
Use the formula, P = I ∆V
I = P / ∆V = 60 / 120 = 0.5 A
∆V = I R
R = ∆V / I = 120 / 0.5 = 240 Ω
Answer:
The kinetic energy of the particle will be 12U₀
Explanation:
Given that,
A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.
Constant force = 12F
According to question,
The kinetic energy is
....(I)
Constant force = 12F
A resistive force field is now set up ,
Resistive force is given by,
When the particle moves from point B to point A then,
We need to calculate the kinetic energy
Using formula for kinetic energy
Put the value of 
Now, from equation (I)
Hence, The kinetic energy of the particle will be 12U₀.
