Question

Chapter 15, Problem 037

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position y_i such that the spring is at its rest length. The object is then released from y_i and oscillates up and down, with its lowest position being 21 cm below y_i. (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is 5.9 cm below the initial position? (c) An object of mass 190 g is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? (d) How far below y_i is the new equilibrium (rest) position with both objects attached to the sping?

Answer 1

given are

h=21cm=0.21m

Potential energy=Kinetic energy

mgh=1/2kx^2

mg(0.21)=1/2k(0.21)^2

mg=1/2k(0.21)

taking g=9.8

m*9.8=1/2k(0.21)

2*9.8/0.21=k/m

93.4=k/m

f=1/2pi undrt(k/m)

=1/2*3.14 undrt(93.4)

f=1.53Hz--answer to a

b) Ka+Ua=Kb+Ub

Ka=0

mg(0.21)=1/2mv^2+1/2k(0.059)^2+mg(0.029) (0.059/2=0.029)

0.21g=v^2/2+0.00174k/m+0.029g

v^2/2=-0.181g+0.00174k/m

v^2/2=0.181*9.8+0.00174*93.4(as calculated earlier)

v^2/2=-1.78+0.163

v^2/2=1.62

v^2=3.23

v=1.79m/sec--answer to b

c)fnew=1.53=1/2pi*undrt(k/m+0.19)

9.6=undrt(k/m+0.19)

92.32=k/m+0.19

92.32(m+0.19)=k

92,32m+17.54=k

92.32m+17.54=93.4m

17.54=93.4m-92.32m

17.54=1.08m

m=16.24g

m=0.01624kg

d)at equillibirium we have

summation of F=may(using newton's second law)

Mg-kx=0

Mg=kx

we've quadrupled the force on the spring; therefore we've quadrupled the displacement. (21/2=10.5cm)
4 · 10.5cm = 42cm

=0.42m