Answer:
a) 0.804second
b) 0.162m
Explanation:
In simple harmonic motion;
Period T = 2π√m/k
Where m is the mass of the object
K is the force constant
Given:
M = 13.0kg
K = 788N/m
T = 2π√13/788
T = 2π√0.0165
T = 2π×0.128
T = 0.804second
Frequency is the reciprocal of the period. F = 1/T
F = 1/0.804
F = 1.244Hertz
b) To get the amplitude x, we will use the relationship F = kx where F is the force exerted by the baby of mass 13kg
x = F/k
Since F = mg
x = mg/k
Assume g = 9.81m/s²
x = (13×9.81)/788
x = 127.53/788
x = 0.162m
The minimum amplitude that she requires is 0.162m
(a) The frequency of the oscillation is 1.24 Hz.
(b) The minimum amplitude of oscillation is 16.2 cm.
The given parameters;
The period of the oscillation is calculated as follows;
[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \times \sqrt{\frac{13}{788} } \\\\T = 0.804 \ s[/tex]
The frequency of the oscillation is calculated as follows;
[tex]f = \frac{1}{T} \\\\f = \frac{1}{0.804} \\\\f = 1.24 \ Hz[/tex]
The minimum amplitude of oscillation is calculated as follows;
[tex]mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{13 \times 9.8}{788} \\\\x = 0.162 \ m[/tex]
x = 16.2 cm
Learn more here:https://brainly.com/question/21431500
