Answer:
The input force (effort) is the amount of effort used to push down on a rod, or pull on a rope in order to move the weight. In this example, the force the little guy is using to pull the elephant is the input force.
Explanation:
Answer:
The frequency of the oscillation is 0.9Hz
Explanation:
This problem bothers on simple harmonic motion of a spring
Given data
Mass of the child m= 25kg
Spring constant k=791 N/m
Amplitude a= 31cm
But the period of the motion as a result of the adults sholve is expressed as
T=2π√m/k
T=2*3.142√25/791
T=6.284√0.031
T=6.284*0.176
T=1.11 sec
But frequency F=1/T
F=1/1.11
F=0.9Hz
Answer:
11.07Hz
Explanation:
Check the attachment for diagram of the standing wave in question.
Formula for calculating the fundamental frequency Fo in strings is V/2L where;
V is the velocity of the wave in string
L is the length of the string which is expressed as a function of its wavelength.
The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)
Therefore L = 1.5λ
If L = 3.0m
1.5λ = 3.0m
λ = 3/1.5
λ = 2m
Also;
V = √T/m where;
T is the tension = 0.98N
m is the mass per unit length = 2.0g = 0.002kg
V = √0.98/0.002
V = √490
V = 22.14m/s
Fo = V/2L (for string)
Fo = 22.14/2(3)
Fo = 22.14/6
Fo = 3.69Hz
Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo
Frequency of the wave = 3×3.69
Frequency of the wave = 11.07Hz