Question

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 7.4 rad/s. The drawing shows the position of the block when the spring is unstrained. This position is labeled ''x = 0 m.'' The drawing also shows a small bottle located 0.080 m to the right of this position. The block is pulled to the right, stretching the spring by 0.050 m, and is then thrown to the left. In order for the block to knock over the bottle,it must be thrown with a speed exceeding v0. Ignoring the width of the block, find v0.

Answer 1

Answer:

v₀ = 0.462 m / s

Explanation:

The spring block system results in an oscillatory movement described by the equation

    x = A cos (wt + φ)

Where A is the amplitude of the movement

Let's analyze the situation presented give the angular velocity, the elongation for t = 0 , and they ask me to hit a bottle that is at x = 0.050 m. The speed is given by

    v = dx / dt

    v = -A w sin (wt + φ)

For the block to hit the bottle the range of motion must be equal to the distance of the bottle

    A = 0.080 m

For t = 0

    x (0) = A cos φ

Let's calculate the phase

    cos φ = x (0) / A

    φ= cos⁻¹ (0.5 / 0.8)

    φi = 0.8957 rad

Let's use the speed equation

    v₀ = -A w sin φ

    v₀ = - 0.080 7.4 sin 0.8957

    v₀ = 0.462 m / s

Answer 2

The speed of the block, in order for the block to knock over the bottle is 0.462 m/s.

Phase angle

The pahse angle of the wave is determined using the following formula;

x = A cosФ

when the position, x = 0.05 m, and maximum displacement = 0.08 m

0.05 = 0.08cosФ

Ф = cos⁻¹(0.05/0.08)

Ф = 0.896 rad

Speed of block

The speed of the block, in order for the block to knock over the bottle is calculated as follows;

v = ωA sin(Φ)

v = 7.4 x 0.08 x sin(0.896)

v = 0.462 m/s

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