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
