Respuesta :
Answer:
a) 250 N/m
b) 22.4 rad/s , 3.6 Hz , 0.28 sec
c) 0.3125 J
Explanation:
a)
F = force applied on the spring = 7.50 N
x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m
k = spring constant of the spring
Since the force applied causes the spring to stretch
F = k x
7.50 = k (0.03)
k = 250 N/m
b)
m = mass of the particle attached to the spring = 0.500 kg
Angular frequency of motion is given as
[tex]w = \sqrt{\frac{k}{m}}[/tex]
[tex]w = \sqrt{\frac{250}{0.5}}[/tex]
[tex]w [/tex] = 22.4 rad/s
[tex]f[/tex] = frequency
Angular frequency is also given as
[tex]w [/tex] = 2 π [tex]f[/tex]
22.4 = 2 (3.14) f
[tex]f[/tex] = 3.6 Hz
[tex]T[/tex] = Time period
Time period is given as
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{3.6}[/tex]
[tex]T[/tex] = 0.28 sec
c)
A = amplitude of motion = 5 cm = 0.05 m
Total energy of the spring-block system is given as
U = (0.5) k A²
U = (0.5) (250) (0.05)²
U = 0.3125 J
(a) The force constant of the spring is 250 N/m.
(b) The angular frequency of the mass oscillation is 22.36 rad/s, frequency is 3.56 Hz and the period is 0.28 s.
(c) the total energy of the system is 0.31 J.
Force constant of the spring
The force constant of the spring can be determined by applying Hooke's law as follows;
F = kx
k = F/x
k = (7.5)/0.03)
k = 250 N/m
Angular frequency
The angular frequency of the mass oscillation is calculated as follows;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{250}{0.5} }\\\\\omega = 22.36 \ rad/s[/tex]
Angular frequency
The angular frequency is calculated as follows;
ω = 2πf
f = ω/2π
f = (22.36)/2π
f = 3.56 Hz
Period of the oscillation
The period of the oscillation is calculated as follows;
T = 1/f
T = 1/3.56
T = 0.28 s
Total energy of the system
The total energy of the system is calculated as follows;
U = ¹/₂kA²
U = ¹/₂ x 250 x (0.05)²
U = 0.31 J
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