Question

A mass of stretches a spring . The mass is set in motion from its equlibrium position with a downward velocity of and no damping is applied. Determine the position of the mass at any time . Use as the acceleration due to gravity. Pay close attention to the units.

Answer 1

Your question is incomplete.

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. u(t) = m b.

Answer:

u(t) = y = 0.0115Cos(8.7t)

Explanation:

Given m = 148g = 0.148kg

x = 13cm = 0.13m = stretched distance

vy = 10cm/s = 0.1m/s = initial downward velocity

g = 9.8m/s²

k = mg/x = 0.148×9.8/0.13 = 11.2N/m = spring constant

ω = √(k/m) = √(11.2/0.148) = 8.7rad/s = angular frequency

A = √(yo² +voy²/ω²) = √(0² +0.1²/8.7²)

A = 0.0115m

y = ACos(ωt) = position with respect to time for a zero phase angle

y = 0.0115Cos(8.7t)

Where A = the amplitude of the motion.

Answer 2

Complete Question:

A mass of 150 g stretches a spring 9 cm. The mass is set in motion from its equilibrium position with a downward velocity of 12 cm/s and no damping is applied. Determine the position of the mass at any time . Use as the acceleration due to gravity. Pay close attention to the units.

Answer:

The position of the mass at any time = speed, $y^{'} = 0.12cos(10.435t)$

Explanation:

Mass, m = 150 g = 150/1000 = 0.15 kg

Extension, x = 9 cm = 0.09 m

According to Hooke's law, F = kx...........(1)

F = mg........(2)

Equating (1) and (2)

mg = kx

$k = \frac{mg}{x}$

$k = \frac{0.15*9.8}{0.09}\\k = 16.33$

The differential equation that models the motion of a spring and a damper

$my^{''} + cy^{'} + ky = 0$

Since there is no damping applied, c = 0

$my^{''} + ky = 0\\0.15y^{''} + 16.33y = 0$

Solving the differential equation above by making $y = e^{mx}$, $y^{'} = me^{mx}$, $y^{''} = m^{2} e^{mx}$

$0.15m^{2} e^{mx} + 16.33m e^{mx} = 0\\0.15m^{2} + 16.33m = 0\\m = \sqrt{\frac{-16.33}{0.15} }$

m = ± 10.435 i

$y = Acos(10.435t) + Bsin(10.435t)$

At the equilibrium position, t = 0, hence y = 0

$0= Acos(10.435*0) + Bsin(10.435*0)$..................(3)

A = 0

Inserting the value of A into equation (3)

$y = Bsin(10.435t)$...........(4)

The position of the mass at any time is equivalent to the speed of the spring

which is the first derivative of y

$y^{'} = 10.435Bcos(10.435t)$..........(5)

Initial speed is given as 12 cm/s = 0.12 m/s, substituting this into equation (5)

$0.12= 10.435Bcos(10.435*0)\\B = 0.12/10.435\\B = 0.0115$

Putting the value of B in equation (5)

$y^{'} = 0.12cos(10.435t)$