Question

1.

a) An apple weighs 1.12 N . When you hang it from the end of a long spring of force constant 1.57 N/m and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back and forth swings do not cause any appreciable change in the length of the spring.)

What is the unstretched length of the spring (i.e., without the apple attached)?

b) On the planet Newtonia, a simple pendulum having a bob with mass 1.50 kg and a length of 165.0 cm takes 1.36 s , when released from rest, to swing through an angle of 12.5 _?, where it again has zero speed. The circumference of Newtonia is measured to be 51400 km.

What is the mass of the planet Newtonia?

Answer 1

Q1.

part a:

frequency in SHM with the spring in vertical direction=(1/(2*pi))*sqrt(k/m)

where k=spring constant=1.57 N/m

m=mass =weight/g=1.12/9.8=0.58989 Hz

when it is swinging from side to side,

frequency=(1/(2*pi))*sqrt(g/L)

given this frequency is half of the bounce frequency

hence (1/(2*pi))*sqrt(g/L)=0.58989

==>sqrt(9.8/L)=3.7064

==>L=0.71338 m

part b:

let radius be r and and mass of the planet be M.

circumference=51400 km=5.14*10^7 m

==>2*pi*r=5.14*10^7

==>r=8.1806*10^6 m

for the pendulum:

time taken to reach from rest to the max angle where it is again at rest is (1/4)th of the time period

hence 1/4*time period=1.36

==>time period=5.44 seconds

as for. a simple pendulum, time period=2*pi*sqrt(L/g)

==>2*pi*sqrt(1.65/g)=5.44

==>g=2.2011 m/s^2

as g for any planet is given as G*M/r^2

==>6.674*10^(-11)*M/(8.1806*10^6)^2=2.2011

==>M=2.2011*(8.1806*10^6)^2/(6.674*10^(-11))

==>M=2.2071*10^24 kg