Question

A 220 g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.150 s. If the total energy of the system is 1.50 J, find the following(a) the force constant of the spring(b) the amplitude of the motion.

Answer 1

a)K=385.86 N/m

b)A=0.088 m

Step-by-step explanation:

Given that

m = 220 gm = 0.22 kg

t= 0.15 s

Total energy TE= 1.5 J

We know that time period in the SHM given as

`T=(2\pi)/(\omega)`

`\omega=(2\pi)/(T)`

`\omega=(2* \pi)/(0.15)`

ω=41.88 rad/s

We know that

ω² m = K

K=Spring constant

`K = 41.88^2 * 0.22 \ N/m`

K=385.86 N/m

`TE=(1)/(2)KA^2`

A=Amplitude

`A=\sqrt{(2* TE)/(K)}`

`A=\sqrt{(2* 1.5)/(385.86)}`

A=0.088 m