Answer: A and B is 1.64 seconds.
The period is the same for the wood and metal pendulums because the mass has been eliminated mathematically in the expression for T.
Explanation: The period T is given by
T = 2π/ω, where ω is the angular frequency.
ω, angular frequency, is represented by
Sqaureroot of (m*g*L/I), where m is mass, L is the distance from pivot to the center of mass and I is the moment of inertia about the pivot .
Therefore
T = 2π/ sqaureroot(m*g*L/I),
Further expressed as
T = 2π*sqaureroot(I/m*g*L),
Inertia Is given by
I=1/3mL^2 where L=length of the stick. Since the meter stick is uniform, the distance L from one end to the center of mass is Upper L = one-half Upper L 0 period.
Thus the period of oscillation is expressed as;
T=2π*sqrt(I/mg(1/2L)
=2π*sqrt(1/3m*L^2/mg(1/2L)
= 2π*sqrt(2L/3g)
=2π*sqrt(2*1/3*9.8)
= 1.64 seconds
A )The Period of the wood pendulum is : 1.638 secs
B) The period of the metal pendulum is ; 1.638 secs
T = 2[tex]\pi[/tex] [tex]\sqrt{(I / ( mgd))}[/tex]
= [tex]2\pi \sqrt{(\frac{(1/3ML^2)}{(MgL/2)} }[/tex]
cancelling out M ( mass ) from the equation
T = [tex]2\pi \sqrt{(2/3*l/g )}[/tex]
= 2π √(2/3 * 1 / 9.81 )
= 1.638 secs
Therefore the period for both wood pendulum and metal pendulum is 1.638 secs because the masses are cancelled out in the equation above.
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