Question

A spring with no mass attached to it hangs from a rigid support. A mass m is now hung on the lower end to the spring. The mass is supported on a platform so that the spring remains relaxed. The supporting platformis then suddenly removed and the mass begins to oscillate. The lowest position of the mass during the oscillation is 5 cm below the place where it was resting on the platform. What is the angular frequency of oscillation? Take g = 10 $m{s}^{-2}$

• A. $10rad{s}^{-1}$
• B. $20rad{s}^{-1}$
• C. $30rad{s}^{-1}$
• D. $40rad{s}^{-1}$

Answer 1

The correct option is B $20rad{s}^{-1}$

It is clear that the separation between the two extreme positions of the oscillating mass is 5 cm. Therefore, the equilibrium position 0 is 2.5 cm below the supporting platform. In other words, force mg produces an extension y = 2.5 cm in the spring. If k is the force constant of the spring we have
m g=k y
or $\frac{k}{m}=\frac{g}{y}=\frac{1000cm{s}^{-2}}{2.5cm}=400s^{-1}$
The angular frequency $\omega$ of oscillation is
$\omega =\sqrt{\frac{k}{m}}=\sqrt{400}=20rad{s}^{-1}$