Question

A block 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| =

Answer 1

Answer:

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

$f=\frac{1}{2\pi} \sqrt{\frac{k}{m}}$

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:

$k=(2\pi f)^2 m=(2 \pi (1.1 Hz))^2 (0.28 kg)=13.4 N/m$

The restoring force of the spring is:

$F=kx$

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:

$x=\frac{F}{k}=\frac{1.0 N}{13.4 N/m}=0.075 m=7.5 cm$