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2003 Q4

679810

Posted 15 January 2007 - 06:33 PM

I bet it's really easy and I'll kick myself for not knowing it but any chance of some help

A mass of 1.5kg is suspended from a spring

This extends the length of the spring from 40mm to 80 mm. The mass is at rest

(a)Calculate the force exerted by the spring on the mass

679810

Posted 15 January 2007 - 07:02 PM

QUOTE(679810 @ Jan 15 2007, 06:33 PM) View Post

I bet it's really easy and I'll kick myself for not knowing it but any chance of some help

A mass of 1.5kg is suspended from a spring

This extends the length of the spring from 40mm to 80 mm. The mass is at rest

(a)Calculate the force exerted by the spring on the mass

Yup I was just looking at the wrong answer so turns out I got it right

Next question:D

(4c)The mass is now pulled down, extending the spring a further 30mm

The mass is released and the subsequent motion is simple harmonic
The force exerted by the spring is directly proportional to its extension

(i)Calculate the unbalanced force acting on the mass just after its release
(ii)Calculate the period of oscillation of the mass

ad absurdum

Posted 16 January 2007 - 06:10 PM

i)The force exerted by the spring is directly proportional to the extension of the spring. So T=kl, where T is the tension in the spring, l is the extension, and k is the constant of proportionality. Use part (a) to find k, then use this value of k to find the tension in the spring for the new extension.

ii)\frac{d^2x}{dt^2}=-\omega ^2 x
\frac{F}{m} = -\omega ^2 a (from part (i), you calculated the force acting at the extremity. Let a be the amplitude of the SHM, and F be the unbalanced force you calculated in (i) - this is just the SHM equation written at the extremity)
-\frac{2\pi}{T} = \sqrt{\frac{F}{mx}}
Solve for T

waitingforan_alibi

Posted 19 January 2007 - 08:41 PM

QUOTE(679810 @ Jan 15 2007, 07:02 PM) View Post
QUOTE(679810 @ Jan 15 2007, 06:33 PM) View Post

I bet it's really easy and I'll kick myself for not knowing it but any chance of some help

A mass of 1.5kg is suspended from a spring

This extends the length of the spring from 40mm to 80 mm. The mass is at rest

(a)Calculate the force exerted by the spring on the mass

Yup I was just looking at the wrong answer so turns out I got it right

Next question:D

(4c)The mass is now pulled down, extending the spring a further 30mm

The mass is released and the subsequent motion is simple harmonic
The force exerted by the spring is directly proportional to its extension

(i)Calculate the unbalanced force acting on the mass just after its release
(ii)Calculate the period of oscillation of the mass



(i)

\begin{align*}F_{up}=kx,  F_{down}=-mg => F_{unb}=kx-mg\end{align*}

(ii) 

\begin{align*}{\omega}^2=\frac{k}{m}\end{align*}


\begin{align*}\omega^2=\frac{4\pi^2}{T^2}\end{align*}

equate and solve for T


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