At the end of the spring of a pendulum with a mass of 0.1 kg, a variable force acts, the frequency of which is 16 Hz.

| May 25, 2021 | education

At the end of the spring of a pendulum with a mass of 0.1 kg, a variable force acts, the frequency of which is 16 Hz. Will there be a resonance in this case if the spring rate is 400 N / m.

m = 0.1 kg.

vc = 16 Hz.

k = 400 N / m.

v -?

The phenomenon of resonance is called a sharp increase in the amplitude of oscillations. For this phenomenon to occur, it is necessary that the oscillation frequency of the external force vс coincides with the frequency of natural free oscillations vс: vс = v.

The frequency of free oscillations of a spring pendulum v is expressed by the formula: v = √k / 2 * P * √m, where k is the stiffness of the spring, P is the number pi, and m is the mass of the load.

v = √400 N / m / 2 * 3.14 * √0.1 kg = 10 Hz.

Since v <vc, the resonance phenomenon will not be observed.

Answer: there will be no resonance.



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