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

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

$T=\dfrac{2\pi}{\omega}$

$\omega=\dfrac{2\pi}{T}$

$\omega=\dfrac{2\times \pi}{0.15}$

ω=41.88 rad/s

We know that

ω² m = K

K=Spring constant

$K = 41.88^2 \times 0.22 \ N/m$

K=385.86 N/m

$TE=\dfrac{1}{2}KA^2$

A=Amplitude

$A=\sqrt{\dfrac{2\times TE}{K}}$

$A=\sqrt{\dfrac{2\times 1.5}{385.86}}$

A=0.088 m