Question

A horizontal spring-mass system has low friction, spring stiffness 205 N/m, and mass 0.6 kg. The system is released with an initial compression of the spring of 13 cm and an initial speed of the mass of 3 m/s.

(a) What is the maximum stretch during the motion? m
(b) What is the maximum speed during the motion? m/s
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?

Answer 1

(a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.

Step-by-step explanation:

Given that,

Spring stiffness = 205 N/m

Mass = 0.6 kg

Compression of spring = 13 cm

Initial speed = 3 m/s

(a). We need to calculate the maximum stretch during the motion

Using conservation of energy

`E_(initial)=E_(final)`

`(1)/(2)kx_(c)^2+(1)/(2)mv^2=(1)/(2)kx_(m)^2`

Put the value into the formula

`(1)/(2)*205*(13*10^(-2))^2+(1)/(2)*0.6*3^2=(1)/(2)*205* x_(m)^2`

`x_(m)=\sqrt{(4.43*2)/(205)}`

`x_(m)=20.7\ cm`

(b). Maximum speed comes when stretch is zero.

We need to calculate the maximum speed during the motion

Using conservation of energy

`E_(initial)=E_(final)`

`(1)/(2)kx_(c)^2+(1)/(2)mv^2=(1)/(2)mv'^2`

Put the value into the formula

`(1)/(2)*205*(13*10^(-2))^2+(1)/(2)*0.6*3^2=(1)/(2)*0.6* v'^2`

`v'=\sqrt{(4.43*2)/(0.6)}`

`v'=3.84\ m/s`

(c). Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system

We need to calculate the time period

Using formula of time period

`T=2\pi\sqrt{(m)/(k)}`

Put the value into the formula

`T=2\pi\sqrt{(0.6)/(205)}`

`T=0.33\ sec`

We need to calculate the energy

Using formula of energy

`E=(P)/(t)`

Put the value into the formula

`E=(0.02)/(0.33)`

`E=0.060\ Watt`

Hence, (a). The maximum stretch during the motion is 20.7 cm

(b). The maximum speed during the motion is 3.84 m/s.

(c). The energy is 0.060 Watt.