HOMEWORK PROBLEMS FOR WEEK # 11PROBLEM 1: Two SPRINGS AND A MASSk2Con si der the sy stem shown in the diagram, which consi st s of two springs withconstants k1 and k2 and a mass m. If all surfaces are frictionless, wh at is theoscillation frequency v of the mass?

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Description: HOMEWORK PROBLEMS FOR WEEK # 11PROBLEM 1: Two SPRINGS AND A MASSk2Con si der the sy stem shown in the diagram, which consi st s of two springs withconstants k1 and k2 and a mass m. If all surfaces are frictionless, wh at is theoscillation frequency

Answered: k1 Con si der the sy stem shown in the…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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HOMEWORK PROBLEMS FOR WEEK # 11
PROBLEM 1: Two SPRINGS AND A MASS
k2
Con si der the sy stem shown in the diagram, which consi st s of two springs with
constants k1 and k2 and a mass m. If all surfaces are frictionless, wh at is the
oscillation frequency v of the mass?

Expert Solution

Step 1

If a mass m is attached to a spring with spring constant k and is displaced from the equilibrium position, a restoring elastic force opposite to the displacement is exerted by the spring.

The equation of motion for such a mass on a spring, is given by using Hooke's law-
$F = m \overset{. .}{x} = - k x$

Step 2

During the oscillation, one spring is stretched and other is compressed and vice-versa.

Let x be the displacement of mass m from its equilibrium position, and this is also the compression and expansion in the springs, $k_{1} a n d k_{2}$ .

The restoring force developed in the two springs is
$F_{1} = - k_{1} x and F_{2} = - k_{2} x$

Since both these forces acts on the mass m in the same direction (towards left in our case of motion of mass). Therefore-
$\begin{array}{rcl} F & = & F_{1} + F_{2} \\ F & = & - k_{1} x - k_{2} x \\ F & = & - (k_{1} + k_{2}) x \end{array}$

or $F = m \overset{. .}{x} = - (k_{1} + k_{2}) x$

Thus the force constant of the system is ( $k_{1} + k_{2}$ )

Since angular frequency is given by: $ω = \sqrt{\frac{k_{system}}{m}}$
$\Rightarrow ω = \sqrt{\frac{k_{1} + k_{2}}{m}} \Rightarrow 2 π ν = \sqrt{\frac{k_{1} + k_{2}}{m}} \Rightarrow ν = \frac{1}{2 π} \sqrt{\frac{k_{1} + k_{2}}{m}}$

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