Question

A mass of 0.26 kg on the end of a spring oscillates with a period of 0.45 s and an amplitude of 0.15 m .a) Find the velocity when it passes the equilibrium point.b) Find the total energy of the system.c) Find the spring constant.d) Find the maximum acceleration of the mass.

Answer 1

Answer:

a)V= 2.09 m/s

b)TE= 0.56 J

c)K=50.47 N/m

d)a=29.23 m/s²

Explanation:

Given that

m = 0.26 kg  ,  T= 0.45 s  ,A= 0.15 m

We know that time period given as

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

ω =Angular frequency

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

${\omega}=\dfrac{2\pi}{0.45}$

ω = 13.96 rad/s

The velocity at equilibrium

V=  ω A

V= 13.96 x 0.15

V= 2.09 m/s

The total energy TE

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

$TE=\dfrac{1}{2}\times 0.26\times 2.09^2$

TE= 0.56 J

The spring constant K

Maximum stored energy in the spring

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

From energy balance

U= TE

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

K A² = m V²

$=\dfrac{mV^2}{A^2}$

$K=\dfrac{0.26\times 2.09^2}{0.15^2}$

K=50.47 N/m

The maximum acceleration a

a= ω² A

a = 13.96²  x 0.15 m/s²

a=29.23 m/s²