A hanging spring stretches by 39.0 cm when an object of mass 460 g is hung on it

ID: 2287908 • Letter: A

Question

A hanging spring stretches by 39.0 cm when an object of mass 460 g is hung on it at rest. We define its position as x = 0. The object is pulled down an additional 17.0 cm and released from rest to oscillate without friction. What is its position x at a moment 84.4 s later? Find the distance traveled by the vibrating object in part (a). (c) Another hanging spring stretches by 39.5 cm when an object of mass 450 g is hung on it at rest. We define this new position as x = 0. This object is also pulled down an additional 17.0 cm and released from rest to oscillate without friction. Find its position x at a time 84.4 s later. (d) Find the distance traveled by the vibrating object in part (c).

Explanation / Answer

spring constant can be find using formula

F=kx

in first case

x=39 cm

F=0.46*9.8

so k=(0.46*9.8)/0.39

w=(k/m)1/2

when a mass is pulled down with to additional distance A=17 cm, it means the particle maximum displacement is 17 cm

part a)

now we know position of particle at any point t

x=Asin(wt)

here t=84.4 s

part b)

time period of vibrating mass

T=2*pi/w

distance covered in this time period is 4A

so when time is T distance is=4A

when time is 84.4 sec distance=(4A/T)*84.4 sec

part c)

x=39.5 cm

F=0.45*9.8

so k=(0.45*9.8)/0.395