Chapter 15, Problem 037 A massless spring hangs from the ceiling with a small ob

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 yi such that the spring is at its rest length. The object is then released from yi and oscillates up and down, with its lowest position being 21 cm below yi. (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 yi is the new equilibrium (rest) position with both objects attached to the sping?

Explanation / Answer

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

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