A 0.78 kg block is attached to a horizontal spring of force constant 256 kg/s2, resting atop a frictionless surface. The block is pulled back 30 cm and then released. What is the period of the oscillations of the block? Maximum speed of the oscillation?

1 Answer

0.347 s0.347s & 5.435\ m/s

Explanation:

The period of oscillation of spring-mass system on any frictionless plane (horizontal, vertical or inclined) is given as follows

T=2\pi\sqrt{m/k}

Where, m mass of system/block & k spring constant of spring

Hence, setting m=0.78 kg & k=256\ text{kg/s}^2 in above formula, the period of oscillation is given as follows

T=2\pi\sqrt{0.78/256}=0.347\ \text{sec}

Now, the maximum speed v_{\text{max}} of block of mass m=0.78 kg will be at mean position when entire elastic energy of spring =1/2kx^2 is transferred to the block when spring is given a displacement x=30 cm

1/2mv_{\text{max}}^2=1/2kx^2

1/2(0.78)v_{\text{max}}^2=1/2(256)(0.3)^2

v_{\text{max}}=5.435\ m/s