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When a \(3-\mathrm{kg}\) block is suspended from a spring, the spring is stretched a distance of \(60 \mathrm{~mm}\). Determine the natural frequency and the period of vibration for a \(0.2-\mathrm{kg}\) block attached to the same spring.

Short Answer

Expert verified
In conclusion, by using Hooke's law to calculate the spring constant, and then applying the formulae for natural frequency and period of vibration, it is possible to find the natural frequency and the period of vibration for a 0.2-kg block attached to the same spring.

Step by step solution

01

Calculation of Spring Constant

Hooke's law states that the force exerted by a spring is proportional to its change in length. So, \(F = k \cdot x\), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the change in length. In this case, \(F\) is the weight of the 3-kg block (\(F = m \cdot g\), where \(m\) is the mass of the block and \(g\) is the acceleration due to gravity, approximately \(9.81 \, m/s^2\)), and \(x\) is 60 mm (0.06 m). Solving the equation for \(k\), you get \(k = F / x\).
02

Determination of Natural Frequency

The natural frequency (\(f\)) of a spring-mass system is given by \(f= (1/2\pi) \cdot \sqrt{k/m}\), where \(k\) is the spring constant (as calculated in Step 1) and \(m\) is the mass of the block. For the 0.2-kg block, \(m = 0.2\) kg. Substituting the known values, calculate the natural frequency.
03

Determination of Period of Vibration

The period of vibration (\(T\)) is the reciprocal of the frequency. Thus, \(T = 1/f\). Once you have calculated the natural frequency, you can calculate the period of vibration by taking its reciprocal.

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