Question

A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the oscillating system is 6.74 J.

A. Determine the following maximum speed of the object?

B. The force constant?

C. The amplitude of the Motion?

Answer 1

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Step-by-step explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating system = 6.74 J

We know that energy is given by

(a)
`Ke=(1)/(2)mv_(max)^2`

`6.74=(1)/(2)* 0.267* v_(max)^2`

`v_(max)=7.1m/sec`

(b) Now
`\omega =(2\pi )/(T)=(2* 3.14)/(0.176)=35.681rad/sec`

We know that
`\omega =\sqrt{(k)/(m)}`

`35.68=\sqrt{(k)/(0.267)}`

`k=339.9N/m`

(c) We know that energy is given by

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

`6.74=(1)/(2)* 339.9* A^2`

`A=19.91cm`