A 130 g block attached to a spring with spring constant 2.8 N/m oscillates horiz

Question

A 130 g block attached to a spring with spring constant 2.8 N/m oscillates horizontally on a frictionless table. Its velocity is 24 cm/s when x0 = -4.4 cm .

Part A

What is the amplitude of oscillation?

Express your answer to two significant figures and include the appropriate units.

Part B

What is the block's maximum acceleration?

Express your answer to two significant figures and include the appropriate units.

Part C

What is the block's position when the acceleration is maximum?

Express your answer to two significant figures and include the appropriate units.

Part D

What is the speed of the block when x1 = 3.3 cm ?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Here ,

let the amplitude of the oscilaltion is A

part A)

Using conservation of energy

0.5 * m * v^2 + 0.5 * k * x0^2 = 0.5 * k * A^2

0.5 * 0.130 * 0.24^2 + 0.5 * 2.8 * 0.044^2 = 0.5 * 2.8 * A^2

solving for A

A = 0.068 m = 6.8 cm

the ampiltude of the motion is 6.8 cm

part B)

as angular frequncy , w = sqrt(k/m)

w = sqrt(2.8/.130)

w = 4.64 rad/s

maximum acceleration , a = A * w^2

maximum acceleration , a = 0.068 * 4.64^2

maximum acceleration , a = 1.464 m/s^2

c)

as the block is at maximum position when the acceleration is maximum

the position of block is -0.068 m ,- 0.068 m

d)

let the speed of the block at x1 = 3.3 cm

Using conservation of energy

0.5 * m * v^2 + 0.5 * k * x1^2 = 0.5 * k * A^2

0.5 * 0.130 * v^2 + 0.5 * 2.8 * 0.033^2 = 0.5 * 2.8 * 0.068^2

v = 0.28 m/s

the speed of the block is 0.28 m/s

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