Question

A 1.40 kg block is attached to a spring with spring constant 14 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 44 cm/s.

What is the amplitude of the subsequent oscillations?

What is the block's speed at the point where x=0.25Ax=0.25A?

Answer 1

Solm, Let amplitude of block is 'A' So, when spring is streched or Compressed by At then block will be at rest, So, using energy conservation equation, Voy initial welocity of block * 1.4 *6.943") vo 44 cm A = 0.1391m TA Angular frequency of block, į kits - Immed fross initial 2 / 1+14* = * 1-4*6-843" 1 = 0.44 smis > = 13.91cm Ano 14 A m >> wo sto rad see So, velocity of block when X= 0.25 A = 0.25 * 0.1391 m = 0.034785 m VE w [A2x2 = són J0.13912 -0.0347852 0.4259 m/s sol vo v= 42.59 cm D Ang see