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 $k$ = 7580 N/m and the mass of the block is $m$ = 0.245 kg

### Solution

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

$F = - k x_m$

Where $k$ 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 $k$ we get

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

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

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

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

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

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

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

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

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