Question

Two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245kg. What is the frequency of oscillation on the frictionless floor?

Solution

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Given

The spring constant is kk = 7580 N/m and the mass of the block is mm = 0.245 kg

Solution

When anybody with mass mm moves under the influence of a Hooke’s law restoring force given by exhibits simple harmonic motion where the force is given by

F=kxmF = - k x_m

Where kk is the spring constant and the negative sign represents the direction of the force. For two springs, we have two times forces, so the force exerted on the block is By solving for kk we get

F=2kxm\begin{equation} F = -2 k x_m \end{equation}

The displacement x(t)x(t) of a particle from its equilibrium position is described by equation 15-3 in the form

x=xmcos(ωt+ϕ)\begin{equation} x=x_{m} \cos (\omega t+\phi) \end{equation}

Where xmx_{m} is the amplitude of the displacement, (ωt+ϕ)(\omega t+\phi) is the phase of the motion, and ϕ\phi is the phase constant. The angular frequency ω\omega is related to the period and frequency of the motion by

ω=2πT=2πf\omega=\frac{2 \pi}{T}=2 \pi f

Take the second derivative for equation (2) we could get the acceleration by

a=ω2xmcos(ωt+ϕ)a=-\omega^{2} x_{m} \cos (\omega t+\phi)

Where the positive term ω2xm\omega^{2} x_{m} represents the maximum acceleration amaxa_{max} when the term cos(ωt+ϕ)=1\cos (\omega t+\phi) = 1.

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