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A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10 cm/s and no damping is applied.
(a) Determine the position u of the mass at any time t. Use 9.8 m/s as the acceleration due to gravity. Pay close attention to the units.
(b) When does the mass first return to its equilibrium position?

Respuesta :

Olajidey

Answer:

[tex]u(t)=1.15 \sin (8.68t)cm[/tex]

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

[tex]\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s[/tex]

Where [tex]g=980 cm/s^2[/tex]

[tex]u(t)=Acos8.68 t+Bsin 8.68t[/tex]

u(0)=0

Substitute the value

[tex]A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t[/tex]

Substitute u'(0)=10

[tex]8.68B=10[/tex]

[tex]B=\frac{10}{8.68}=1.15[/tex]

Substitute the values

[tex]u(t)=1.15 \sin (8.68t)cm[/tex]

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

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