A0.650 kg block is attached to a spring with spring constant 18.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 47.0 cm/s . what are
the amplitude of the subsequent oscillations? answer is in cm
the block's speed at the point where x= 0.350 a? answer is in cm/s

amplitude is 8.92 cm

speed of block is 44.11 cm/s

Explanation:

given data

mass = 0.650 kg

spring constant = 18 N/m

speed = 47 cm/s = 0.44 m/s

speed at x point = 0.350 A

to find out

amplitude of subsequent oscillation

solution

we know here conservation of energy

maximum kinetic energy = maximum potential energy

here we know k is spring constant and m is mass and A is amplitude and v is velocity

so solve it we get

A =

put here all these value

A =  

A = 0.08931 m

so amplitude is 8.92 cm

and

by conservation of energy

initial energy = final energy

=

solve we gey V

V =    

put here value Vm = 0.47 , and x = 0.350

V =

V = 0.4411 m/s

so speed of block is 44.11 cm/s