Question

A 0.650 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

Answer 1

Answer:

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

$\frac{1}{2} m v^{2} = \frac{1}{2} k A^{2}$

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 = $\sqrt{\frac{mv^{2} }{k} }$

put here all these value

A =  $\sqrt{\frac{0.650(0.47)^{2} }{18} }$

A = 0.08931 m

so amplitude is 8.92 cm

and

by conservation of energy

initial energy = final energy

$\frac{1}{2} m V^{2} + \frac{1}{2} k x^{2}$ = $\frac{1}{2} m Vm^{2} + 0$

solve we gey V

V = $\sqrt{Vm^{2} - \frac{k }{m} x^{2} }$    

put here value Vm = 0.47 , and x = 0.350

V = $\sqrt{0.47^{2} - \frac{18 }{0.650} (0.350*0.088 m)^{2} }$

V = 0.4411 m/s

so speed of block is 44.11 cm/s