1. 1.59 s
The period of a pendulum is given by:
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]
where L is the length of the pendulum and g the gravitational acceleration.
In this problem,
L = 0.625 m
g = 9.81 m/s^2
Substituting into the equation, we find
[tex]T=2\pi \sqrt{\frac{(0.625 m)}{9.81 m/s^2}}=1.59 s[/tex]
2. 54,340 oscillations
The total number of seconds in a day is given by:
[tex]t=24 h \cdot 60 min/h \cdot 60 s/min =86,400 s[/tex]
So in order to find the number of oscillations of the pendulum in one day, we just need to divide the total number of seconds per day by the period of one oscillation:
[tex]N=\frac{t}{T}=\frac{86,400 s}{1.59 s}=54,340[/tex]
3. 0.842 m
We want to increase the period of the pendulum by 16%, so the new period must be
[tex]T'=T+0.16T=1.16 T = 1.16 (1.59 s)=1.84 s[/tex]
Now we can re-arrange the equation for the period of the pendulum, using T=1.84 s, to find the new length of the pendulum that is required to produce this value of the period:
[tex]L=g(\frac{T}{2\pi})^2=(9.81 m/s^2)(\frac{1.84 s}{2\pi})^2=0.842 m[/tex]
