Question

A 1.35 kgkg block is attached to a spring with spring constant 15.0 N/mN/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 34.0 cm/scm/s . What are

Answer 1

Given :

Mass of block , m = 1.35 kg .

Speed constant , k = 15 N/m .

Speed at centre , v = 34 cm/s = 0.34 m/s .

To Find :

The amplitude of oscillation .

Solution :

Now , we know total energy is conserved in SHM .

So , K.E at zero displacement :

`K.E=(mv^2)/(2)\\\\K.E=(1.35* 0.34^2)/(2)\\\\K.E=7.8* 10^(-2)\ J`

Now , this K.E is equal to maximum P.E :

`P.E=K.E=7.8* 10^(-2)\ J` .

`P.E=(kA^2)/(2)\\\\7.8* 10^(-2)=(kA^2)/(2)\\\\A=\sqrt{(2* 7.8 * 10^(-2))/(15)}\ m\\\\A=0.1019 \ m\\\\A=10.2\ cm`

Therefore , the amplitude is 10.2 cm .

Hence , this is the required solution .