Answer:
1) [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex], 2) [tex]T_{B} \approx 1.137\cdot T_{A}[/tex], where [tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex].
Explanation:
1) Pendulum A is a simple pendulum, whose period ([tex]T_{A}[/tex]) is determined by the following formula:
[tex]T_{A} = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex] (1)
Where:
[tex]l[/tex] - Length of the massless bar.
[tex]g[/tex] - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period ([tex]T_{B}[/tex]) is determined by the following formula:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{I_{O}}{m\cdot g\cdot l} }[/tex] (2)
Where:
[tex]m[/tex] - Total mass of the pendulum.
[tex]g[/tex] - Gravitational acceleration.
[tex]l[/tex] - Length of the uniform bar.
[tex]I_{O}[/tex] - Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_{O} = \frac{1}{2} \cdot m\cdot l^{2}+\frac{1}{24}\cdot m\cdot l^{2} + \frac{3}{4}\cdot m\cdot l^{2}[/tex]
[tex]I_{O} = \frac{31}{24}\cdot m\cdot l^{2}[/tex] (3)
By applying (3) in (2) we get the following expression:
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{\frac{31}{24}\cdot m \cdot l^{2} }{m\cdot g \cdot l} }[/tex]
[tex]T_{B} = 2\pi \cdot \sqrt{\frac{31\cdot l}{24\cdot g} }[/tex]
[tex]T_{B} = \sqrt{\frac{31}{24} } \cdot \left(2\pi \cdot \sqrt{\frac{l}{g} }\right)[/tex]
[tex]T_{B} \approx 1.137\cdot T_{A}[/tex]
1. The period of pendulum A for small oscillations is
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
2. The period of pendulum B for small oscillations.
[tex]T_B=1.137.T_A[/tex]
Simple harmonic motion is the periodic motion or back and forth motion of any object with respect to its equilibrium or mean position. The restoring force is always acting on the object which try to bring it to the equilibrium.
1) Pendulum A is a simple pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{L}{g}}[/tex]
Where:
l - Length of the massless bar.
g - Gravitational acceleration.
2) Pendulum B is a physical pendulum, whose period () is determined by the following formula:
[tex]T_A=2\pi\sqrt{\dfrac{I_o}{mgl}}[/tex] .............................2
Where:
m - Total mass of the pendulum.
g - Gravitational acceleration.
l - Length of the uniform bar.
Io- Moment of inertia of the pendulum with respect to its suspension axis.
The moment of inertia can be found by applying the formulae of the moment of inertia for a particle and the uniform bar and Steiner's Theorem:
[tex]I_o=\dfrac{1}{2}ml^2+\dfrac{1}{24}ml^2+\dfrac{3}{4}ml^2[/tex]
[tex]I_o=\dfrac{31}{24}ml^2[/tex]..................................3
By applying (3) in (2) we get the following expression:
[tex]T_B=2\pi\sqrt{\dfrac{\frac{31}{24}ml^2}{mgl}[/tex]
[tex]T_B=2\pi\sqrt{\dfrac{31l}{24g}}[/tex]
[tex]T_B=\sqrt{\dfrac{31}{24}}. (2\pi\sqrt{\dfrac{l}{g}})[/tex]
[tex]TB=1.137.T_A[/tex]
Thus to know more about Simple harmomnic motion follow
https://brainly.com/question/17315536
