Respuesta :
Answer:
a)
[tex]135[/tex]Nm⁻¹
b)
[tex]0.925[/tex] Hz
c)
[tex]1.2[/tex]ms⁻¹
d)
[tex]0[/tex] m
e)
[tex]6.7[/tex]ms⁻²
f)
[tex]\pm 0.2[/tex] m
Explanation:
a)
[tex]F[/tex] = force required to hold the object at rest connected with stretched spring = 27 N
[tex]x[/tex] = stretch in the spring from equilibrium position = 0.2 m
[tex]k[/tex] = force constant of the spring
force required to hold the object at rest is same as the spring force , hence
[tex]F = k x[/tex]
[tex]k = \frac{F}{x}[/tex]
inserting the values
[tex]k = \frac{27}{0.2} = 135[/tex] Nm⁻¹
b)
frequency of the oscillations is given as
[tex]f =\frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]
inserting the values
[tex]f =\frac{1}{2(3.14) }\sqrt{\frac{135}{4}}\\f = 0.925[/tex] Hz
c)
[tex]A[/tex] = Amplitude of oscillations = 0.2 m
[tex]w[/tex] = angular frequency
Angular frequency is given as
[tex]w = 2\pi f = 2 (3.14) (0.925) = 5.8[/tex] rads⁻¹
Maximum speed of oscillation is given as
[tex]v_{max} = Aw[/tex]
[tex]v_{max} = (0.2)(5.8)\\v_{max} = 1.2[/tex] ms⁻¹
d)
maximum speed of the object occurs at the equilibrium position, hence
[tex]x = 0[/tex] m
e)
Maximum acceleration of oscillation is given as
[tex]a_{max} = Aw^{2}[/tex]
[tex]a_{max} = (0.2)(5.8)^{2}\\a_{max} = 6.7[/tex]ms⁻²
f)
maximum acceleration occurs when the object is at extreme positions, hence
[tex]x = \pm 0.2[/tex] m
