A spring of force constant 1200 Nm^(-1) is mounted on a horizontal table as shown in Fig. A mass of 3.0 kg is attached to the free end of the spring, pulled side ways to a distance 2.0 cm and released. Determine the maximum acceleration of the mass.

A spring of force constant 1200 Nm^(-1) is mounted on a horizontal tab

1.	A spring of force constant 1200 Nm^(-1) is mounted on a horizontal table as shown in Fig. A mass of 3.0 kg is attached to the free end of the spring, pulled side ways to a distance 2.0 cm and released. Determine the maximum acceleration of the mass.
Answer» Solution :Here, `k=1200 NM^(-1) , m= 3.0 kg , A = 2.0 CM = 0.02 m` (a) Frequency , `f= 1/T = 1/(2A) sqrt(k/m) = 1/(2xx3.14) sqrt((1200)/(3)) = 3.2 s^(-s)` Acceleration `a= omega^2 y = k/my ` Accerlation will be maximum when y is maximum i.e. y = A `therefore ` Max. acceleration , `a_("max") = (kA)/(m) = (1200xx0.02)/(3) = 8 ms^(-2)`