Question

Find the natural frequency of the system shown in the figure. The pulleys are smooth and massless.

Answer 1

Hint : The springs of the system can be considered to be equivalent to two springs in parallel. The tension in each of the strings can be said to lift a quarter of the mass of the body attached to the pulley.

Formula used: In this solution we will be using the following formula;
${k_{eq}} = {k_1} + {k_2}$ where ${k_{eq}}$ is equivalent spring constant of two springs connected in parallel, and ${k_1}$ and ${k_2}$ are the spring constant of the individual springs.
$\omega = \sqrt {\dfrac{k}{m}}$ where $\omega$ is the natural frequency of a system of one spring attached to a mass.

Complete step by step answer:

Let us assume that a mass $M$ was attached to the system as shown. At equilibrium, the tension in the springs and the strings ${T_1}...{T_4}$ are equal. Hence, ${T_1}...{T_4} = T$
Now, the equivalent spring constant of the system can be given as
${K_{eq}} = K + K = 2K$
This is because, for two springs in parallel (i.e. independently attached to the same mass), the equivalent spring constant of the system is
${k_{eq}} = {k_1} + {k_2}$ where ${k_{eq}}$ is equivalent spring constant of two springs connected in parallel, and ${k_1}$ and ${k_2}$ are the spring constant of the individual springs.
Now, since a total tension of ${T_1} + {T_2} + {T_3} + {T_4} = T + T + T + T = 4T$ , then the mass carried by each spring is actually $\dfrac{M}{4}$ .
These two implications signify that a spring is as though it has a spring constant of $2K$ , and carries only a mass of $\dfrac{M}{4}$ .
Now, $\omega = \sqrt {\dfrac{k}{m}}$ where $\omega$ is the natural frequency of a system of one spring attached to a mass.
Hence, for the above, the natural frequency of the system would be
$\omega = \sqrt {\dfrac{{2K}}{{\dfrac{M}{4}}} = } \sqrt {2K \times \dfrac{4}{M}}$
Hence,
$\omega = \sqrt {\dfrac{{8K}}{M}}$ .

Note:
For clarity, the natural frequency of a body is the frequency at which it will oscillate if it were displaced from equilibrium. Hence, a system does not only have a natural frequency only when it is oscillating. The existence of the system itself creates a natural frequency.
An application of it is to create or avoid resonance. Engineers calculate the natural frequencies of buildings, bridges, etc to ensure that nothing in the vicinity will allow the building to be set in resonance and will cause demolition of the building or bridge.