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
Harish Chandra Rajpoot
Jul 4, 2018

#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#

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