A 0.1 kg mass hanging from a light helical spring produces an equilibrium extension of 0.1 m.The mass is pulled vertically downwards by a distance of 0.02m and then released.

Taking g =10 N/kg, the equation relating displacement x of the mass from its equilibrium position and the time t after release is: *(1 Point) \frac{x}{m}=0.02 \cos \left[10\left(\frac{t}{s}\right)\right] \frac{x}{m}=0.02 \cos \left[0.1\left(\frac{t}{s}\right)\right] \frac{x}{m}=0.1 \sin \left[10\left(\frac{t}{s}\right)\right] \frac{x}{m}=0.02 \sin \left[0.2 \pi\left(\frac{t}{s}\right)\right] \frac{x}{m}=0.1 \cos \left[0.2 \pi\left(\frac{t}{s}\right)\right]