Question

A 0.150 kg toy is undergoing SHM on the end of a horizontal spring with force constant 300.0 N/m. When the object is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.200 m/s. Find(a) the total energy of the object at any point in its motion,(b) the amplitude of the motion, and(c) the maximum speed attained by the object during its motion.

Answer 1

a)TE=0.0245 J

b)A = 0.0128 m

c)V=0.57 m/s

Step-by-step explanation:

Given that

m = 0.150 kg

K= 300 N/m

x= 0.012 ,v= 0.2 m/s

The velocity of the toy at any point given as

`v=\omega√(A^2-x^2)`

`\omega=\sqrt{(K)/(m)}`

`v=\sqrt{(K)/(m)}* √(A^2-x^2)`

`0.2=\sqrt{(300)/(0.15)}* √(A^2-0.012^2)`

2 x 10⁻⁵ = A² - 0.000144

A=0.0128 m

Amplitude ,A = 0.0128 m

The total energy TE

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

`TE=(1)/(2)300* 0.0128^2`

TE=0.0245 J

The maximum speed

`V=\omega A`

`V=\sqrt{(K)/(m)}* A`

`V=\sqrt{(300)/(0.15)}* 0.0128`

V=0.57 m/s