Respuesta :
Answer:
a) [tex]v_{max}=0.4\ m.s^{-1}[/tex]
b) [tex]a_{max}=1.6\ m.s^{-2}[/tex]
c) [tex]v_x=0.32\ m.s^{-1}[/tex]
d) [tex]a_x=0.96\ m.s^{-1}[/tex]
e) [tex]\Delta t=0.232\ s[/tex]
Explanation:
Given:
mass of the object attached to the spring, [tex]m=0.5\ kg[/tex]
spring constant of the given spring, [tex]k=8\ N.m^{-1}[/tex]
amplitude of vibration, [tex]A=0.1\ m[/tex]
a)
Now, maximum velocity is obtained at the maximum Kinetic energy and the maximum kinetic energy is obtained when the whole spring potential energy is transformed.
Max. spring potential energy:
[tex]PE_s=\frac{1}{2} .k.A^2[/tex]
[tex]PE_s=0.5\times 8\times 0.1^2[/tex]
[tex]PE_s=0.04\ J[/tex]
When this whole spring potential is converted into kinetic energy:
[tex]KE_{max}=0.04\ J[/tex]
[tex]\frac{1}{2}.m.v_{max}^2=0.04[/tex]
[tex]0.5\times 0.5\times v_{max}^2=0.04[/tex]
[tex]v_{max}=0.4\ m.s^{-1}[/tex]
b)
Max. Force of spring on the mass:
[tex]F_{max}=k.A[/tex]
[tex]F_{max}=8\times 0.1[/tex]
[tex]F_{max}=0.8\ N[/tex]
Now acceleration:
[tex]a_{max}=\frac{F_{max}}{m}[/tex]
[tex]a_{max}=\frac{0.8}{0.5}[/tex]
[tex]a_{max}=1.6\ m.s^{-2}[/tex]
c)
Kinetic energy when the displacement is, [tex]\Delta x=0.06\ m[/tex]:
[tex]KE_x=PE_s-PE_x[/tex]
[tex]\frac{1}{2} .m.v_x^2=PE_s-\frac{1}{2} .k.\Delta x^2[/tex]
[tex]\frac{1}{2}\times 0.5\times v_x^2=0.04-\frac{1}{2} \times 8\times 0.06^2[/tex]
[tex]v_x=0.32\ m.s^{-1}[/tex]
d)
Spring force on the mass at the given position, [tex]\Delta x=0.06\ m[/tex]:
[tex]F=k.\Delta x[/tex]
[tex]F=8\times 0.06[/tex]
[tex]F=0.48\ N[/tex]
therefore acceleration:
[tex]a_x=\frac{F}{m}[/tex]
[tex]a_x=\frac{0.48}{0.5}[/tex]
[tex]a_x=0.96\ m.s^{-1}[/tex]
e)
Frequency of oscillation:
[tex]\omega=\sqrt{\frac{k}{m} }[/tex]
[tex]\omega=\sqrt{\frac{8}{0.5} }[/tex]
[tex]\omega=4\ rad.s^{-1}[/tex]
So the wave equation is:
[tex]x=A.\sin\ (\omega.t)[/tex]
where x = position of the oscillating mass
put x=0
[tex]0=0.1\times \sin\ (4t)[/tex]
[tex]t=0\ s[/tex]
Now put x=0.08
[tex]0.08=0.1\times \sin\ (4t)[/tex]
[tex]t=0.232\ s[/tex]
So, the time taken in going from point x = 0 cm to x = 8 cm is:
[tex]\Delta t=0.232\ s[/tex]
