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.

Respuesta :

Answer:

a)K=385.86 N/m

b)A=0.088 m

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

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]\omega=\dfrac{2\pi}{T}[/tex]

[tex]\omega=\dfrac{2\times \pi}{0.15}[/tex]

ω=41.88 rad/s

We know that

ω² m = K

K=Spring constant

[tex]K = 41.88^2 \times 0.22 \ N/m[/tex]

K=385.86 N/m

[tex]TE=\dfrac{1}{2}KA^2[/tex]

A=Amplitude

[tex]A=\sqrt{\dfrac{2\times TE}{K}}[/tex]

[tex]A=\sqrt{\dfrac{2\times 1.5}{385.86}}[/tex]

A=0.088 m