Question

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?

Answer 1

`0.347 s` & `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`