Question

A simple harmonic oscillator consists of a 0.96 kg block attached to a spring (k = 220 N/m). The block slides on a horizontal frictionless surface about the equilibrium point x = 0 with a total mechanical energy of 3.0 J. (a) What is the amplitude of the oscillation? (b) How many oscillations does the block complete in 12 s? Enter the integer number of complete cycles. (c) What is the maximum kinetic energy attained by the block? (d) What is the speed of the block at x = 0.12 m?

Answer 1

a)
`A=0.165m`

b) 29

c)
`K.E_(max)=3.0\ \textup J`

d)
`V=1.713m/s`

Step-by-step explanation:

mass, m =0.96 g

k = 220 N/m

Total energy, E = 3.0 J

Now,

a)
`E=(1)/(2)kA^2`

where, A is the amplitude

on substituting the values, we get

`3.0=(1)/(2)* 220* A^2`

or

`A^2=(3*2)/(220)`

or

`A^2=0.02727`

or

`A=0.165m`

b) Time period (T) is given as:

`T=2\pi\sqrt(m)/(k)`

on substituting the values,we get

`T=2\pi\sqrt(0.96)/(220)`

or

`T=0.415s`

thus, number of oscillations (N) in 12s will be

`N=(12)/(0.415)=28.91\approx29\ \textup{oscillations}`

c)Maximum K.E = total mechanical energy

thus,

`K.E_(max)=3.0\ \textup J`

d)The angular frequency (ω) is given as:

`\omega=\sqrt(k)/(m)`

on substituting the values,we get

`\omega=\sqrt(220)/(0.96)`

or

`\omega=15.13 s^(-1)`

Now, the speed (V) in SHM is calculated as;

`V=\omega√(A^2-x^2)`

for x = 0.12m

we get

`V=15.13* √(0.165^2-0.12^2)`

or

`V=1.713m/s`