Ablock with a mass of 0.28 kg is attached to a horizontal spring. the block is pulled back from its equilibrium position until the spring exerts a force of 1.0 n on the block. when the block is released, it oscillates with a frequency of 1.1 hz. how far was the block pulled back before being released? express your answer with the appropriate units. |δx| =

The block was pulled at a distance of 0.0823m before being released

First, we need to get the spring constant using the formula for the frequency of the spring as shown;

Given the following parameters

mass m = 0.28kg

F = 1.0N

f = 1.1Hz

Substitute the given parameters into the formula to have:

Next is to get the distance that the block is being pulled before release. According to Hooke's law;

F = kx

x = F/k

x = 1.1/13.3616

x = 0.0823m

Hence the block was pulled at a distance of 0.0823m before being released

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0.075 m = 7.5 cm

Explanation:

In a simple harmonic motion, the frequency of oscillation (f) is related to the mass (m) and the spring constant (k) by the formula

In this problem, we know

f = 1.1 Hz

m = 0.28 kg

So we can re-arrange this formula to find the spring constant:

The restoring force of the spring is:

where

F = 1.0 N is the force exerted on the block

x is the displacement of the block

Therefore, by re-arranging this equation we can find how far was the block pulled back before being released:

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