Frequency of Oscillation for a Spring with Attached Block

AI Thread Summary
To find the frequency of oscillation for a spring with an attached block, the equilibrium position is established when the weight of the block equals the restoring force of the spring. The relevant equations include the frequency formula f = ω/2π and the angular frequency ω = √(k/m). The discussion clarifies that the restoring force can be expressed as F = kx, leading to the relationship k/m = g/x, where g is the acceleration due to gravity and x is the stretch of the spring. This allows for the calculation of the frequency once k/m is determined. Understanding these relationships is crucial for solving the problem accurately.
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Homework Statement



A spring is hung from the ceiling. When a block is attached to its end, it streches 19.3 cm before reaching its new equilibrim length. The block is then pulled down slightly and released. What is the frequency of the oscillation?

Homework Equations


f =\frac{\omega}{2\pi}

\omega= \sqrt{\frac{k}{m}}

The Attempt at a Solution


i have no idea how to start this question and what to make of the mention of the equilibrium length pls. help
 
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Which two opposite and equal forces act upon the mass when the spring is in its equilibrium position?
 
the weight and the normal force but they don't mention any mass in the question
 
You don't want to calculate the mass you want to calculate k/m. The restoring force exerted by the spring is equal to the weight. Can you write this down in formulaic form?
 
wld it be .5*k*x^2= mg where K/m = 2g/x^2 ??
 
That is not Hooke's law. You used the elastic potential energy as a force which is not true because it is an energy. Find the correct expression for the restoring force exerted by the spring.
 
oh do u mean F=kx so that wld be kx=mg where k/m = g/x
 
Yep.
 
thanx i get it now :)
 
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