Question

(III) A block of mass m is supported by two identical parallel vertical springs, each with spring stiffness constant k (Fig. 11–53). What will be the frequency of vertical oscillation?

Short answer

The frequency of vertical oscillation will be \(\frac{1}{{2\pi }}\sqrt {\frac{{2k}}{m}} \).

Step-by-step solution

Understanding simple harmonic motion

Any vibrating object undergoes simple harmonic motion ifthe restoring force (F) developed in the object is proportional to the displacement (x), that is,

\(F = - kx\)

The number of cycles of oscillation completed in one second is referred as the frequency of oscillation. The frequency of oscillation of a mass m attached to the end of a spring of spring constant k is given as:

\(f = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} \)

Evaluation of forces acting on the block in equilibrium position

Let us say that the spring on the left side is A and the spring on the right side is B.

When the block is in equilibrium position, various forces acting on it are:

(i) Spring force due to spring A \(\left( {{{\vec F}_{\rm{A}}}} \right)\), acting upwards

(ii) Spring force due to spring B \(\left( {{{\vec F}_{\rm{B}}}} \right)\), acting upwards

(iii) Weight of the block (mg), acting downwards

On applying Newton’s second law in the vertical direction, considering upward direction to be the positive direction, you will get:

\(\begin{aligned}{c}\sum {{F_y}} &= 0\\{F_{\rm{A}}} + {F_{\rm{B}}} - mg &= 0\\{F_{\rm{A}}} + {F_{\rm{B}}} &= mg\end{aligned}\) ..... (i)

Here, g is the gravitational acceleration.

Evaluation of forces acting on the block in the displaced position

When the block is displaced from its equilibrium position through a distance of x in the downward direction, then the spring force exerted by each spring will get increase by the amount -kx.

Thus, the spring forces exerted by the spring A and spring B become:

\({F_A}^\prime = {F_A} - kx\)

\({F_{\rm{B}}}^\prime = {F_{\rm{B}}} - kx\)

So, again on applying the Newton’s second law in the vertical direction, you will get:

\(\begin{aligned}{c}\sum {{F_y}} &= {F_{\rm{A}}}^\prime + {F_{\rm{B}}}^\prime - mg\\ &= {F_{\rm{A}}} - kx + {F_{\rm{B}}} - kx - mg\\ &= {F_{\rm{A}}} + {F_{\rm{B}}} - 2kx - mg\end{aligned}\) ….. (ii)

Determination of frequency of vertical oscillation

Using (i) in the expression (ii), you will get:

\(\begin{aligned}{c}\sum {{F_y}} &= mg - 2kx - mg\\ &= - 2kx\end{aligned}\)

Thus, the value of effective spring constant is, \({k_{effective}} = 2k\).

So, the frequency of vertical oscillation will be:

\(\begin{aligned}{c}f &= \frac{1}{{2\pi }}\sqrt {\frac{{{k_{{\rm{effective }}}}}}{m}} \\f &= \frac{1}{{2\pi }}\sqrt {\frac{{2k}}{m}} \end{aligned}\)

Thus, the frequency of vibration is \(f = \frac{1}{{2\pi }}\sqrt {\frac{{2k}}{m}} \).