Question

A 2.00-kg frictionless block attached to an ideal spring with force constant 315 N/m is undergoing simple harmonic motion. When the block has displacement +0.200 m, it is moving in the negative x-direction with a speed of 4.00 m/s.
Find
(a) the amplitude of the motion;
(b) the block’s maximum acceleration; and
(c) the maximum force the spring exerts on the block.

Answer 1

Answer:

Explanation

The answer is given in the attachment

Answer 2

The motion of the frictionless block can be described as a simple harmonic motion SHM.

• (a) The amplitude of the motion is approximately 0.376 m

• (b) The maximum acceleration is approximately 59.26 m/s²

• (c) The maximum force the spring exerts on the block is approximately 118.44 N

Reasons:

(a) The displacement of a spring in the x-direction is given by the equation;

v = ω·√(A² - x²)

$\omega = \sqrt{\dfrac{K}{m} } = \sqrt{\dfrac{315 \, N/m}{2.00 \ kg} } \approx 12.55 \ rad/s$

4.00 ≈ 12.55 × √(A² - 0.2²)

Solving gives;

$A =\sqrt{ \left(\dfrac{4.00}{12.55} \right)^2+ 0.2^2} \approx 0.376$

The amplitude, A ≈ 0.376 m.

(b) The force, F = -K·x

F = m·a

$The \ acceleration, \ a = \mathbf{\dfrac{-K \cdot x}{m}}$

K, and m are constant, therefore. the maximum acceleration is given by

the maximum x = The amplitude, A

Therefore, we get;

$The \ magnitude \ maximum \ acceleration, \ a = \dfrac{315 \times 0.376}{2.0} \approx 59.26$

The maximum acceleration, a ≈ 59.26 m/s².

(c) Force, F = Mass, m × Acceleration, a

Therefore, the maximum force occurs at the point where the acceleration is

maximum, which is at the point, x = A

For the spring, F = -K·x

Therefore, $F_{max}$ = -K × A ≈ -315 N/m × 0.376 m = 118.44 N

The maximum force the spring exerts on the block, $F_{max}$ ≈ 118.44 N.

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