Answer:
42.5 Hz.
Explanation:
The fundamental frequency of a closed pipe is given as
f₀ = v/4l....................... Equation 1
Where f₀ = lowest frequency, v = speed of sound in air, l = length of the organ pipe
Given: v = 340 m/s, l = 2.00 m.
Substitute into equation 1
f₀ = 340/(4×2)
f₀ = 340/8
f₀ = 42.5 Hz.
Hence the smallest frequency that will resonant in the organ pipe = 42.5 Hz.
Explanation:
Given that,
Mass of disk = 1.2 kg
Radius = 0.07 m
Radius of rod = 0.11 m
Mass of small disk = 0.5 kg
Force = 29 N
Time t = 0.022 s

Distance d= 0.039 m
(I). We need to calculate the speed of the apparatus
Using work energy theorem



Where, m = total mass
v = velocity
F = force
d = distance
Put the value into the formula


(b). We need to calculate the angular speed of the apparatus
Using formula of torque





We need to calculate the angular speed of the apparatus
Using equation of angular motion

Put the value into the formula


(c). We need to calculate the angular speed of the apparatus
Using equation of angular motion

Put the value into the formula


Hence, This is required equation.
Solution:
f ( t )= 20 S ( t ) + 55/30 tS ( t )− 55/30 ( t − 30 ) S ( t − 30 )
• Taking the Laplace Transform:
F ( s ) = 20/s + 55/30 ( 1/s^2 ) – 55/30 ( 1/s^2) e^-30s = 20/s + 55/30 ( 1/s^2 ) ( 1 – e^-30s)