What Is the Frequency of Oscillation for a Mass on a Spring?

  • Thread starter Thread starter sharkasm
  • Start date Start date
  • Tags Tags
    Frequency Spring
AI Thread Summary
The discussion revolves around calculating the frequency of oscillation for a mass attached to a spring. The problem states that the spring is massless and the object oscillates 10 cm below its initial position. Participants emphasize the need to correctly apply the equations of motion, particularly the relationship between spring constant (k), mass (m), and frequency (f). Key points include the importance of identifying the equilibrium position and the necessity of knowing both k and m to solve for frequency. The conversation highlights the challenges faced due to missing information about the spring constant and mass.
sharkasm
Messages
4
Reaction score
0

Homework Statement


This is number 23 in ch. 15 of Halliday, Resnick, and Walker 's Fundamentals of Physics.
" A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at a rest position y_i such that the spring is at its rest length. The object is then released from y_i and oscillates up and down, with its lowest position being 10 cm below y_i. What is the frequency of the oscillation?


Homework Equations


<br /> F=-kx, \, y&#039;&#039;+\frac{k}{m}y=-g, y=Acos(\sqrt{\frac{k}{m}}+\phi)+\frac{-gm}{k}<br /> [tex\]<br /> mgh+\frac{1}{2}mv^2=\frac{1}{2}kx^2 [\tex]&lt;br /&gt; f=2\pi\omega=2\pi\sqrt{\frac{k}{m}}[\tex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; &amp;lt;h2&amp;gt;The Attempt at a Solution&amp;lt;/h2&amp;gt;&amp;lt;br /&amp;gt; I just solved the differential equation, i can&amp;amp;#039;t seem to relate that to find the frequency, unless its something simple I&amp;amp;#039;ve overlooked. It seems like I don&amp;amp;#039;t have enough information to solve it...
 
Physics news on Phys.org
sharkasm said:

Homework Equations


<br /> F=-kx, \, y&#039;&#039;+\frac{k}{m}y=-g, y=Acos(\sqrt{\frac{k}{m}}+\phi)+\frac{-gm}{k}<br /> [tex\]<br /> mgh+\frac{1}{2}mv^2=\frac{1}{2}kx^2 [\tex]&lt;br /&gt; f=2\pi\omega=2\pi\sqrt{\frac{k}{m}}[\tex]&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; &amp;lt;h2&amp;gt;The Attempt at a Solution&amp;lt;/h2&amp;gt;&amp;lt;br /&amp;gt; I just solved the differential equation, i can&amp;amp;#039;t seem to relate that to find the frequency, unless its something simple I&amp;amp;#039;ve overlooked. It seems like I don&amp;amp;#039;t have enough information to solve it...
&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Check the equations, if you copied them correctly. &amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; What is the meaning of f in the third equation? &amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; ehild
 
i forget a t next to \sqrt{\frac{k}{m}}[\tex] in the solution for y<br /> f is the frequency of the oscillator
 
The equation for the frequency is also wrong. Check it.

Neither the spring constant nor the mass of the object was given. But you should know that the object oscillates around the equilibrium position where it would stay if you released it very slowly. At this equilibrium position the resultant force on the object is zero. Write this condition, and you find k/m. ehild
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top