Frequency of a simple pendulum and it's mass

AI Thread Summary
The frequency of a simple pendulum is primarily affected by its length and amplitude, while mass does not influence the frequency due to gravity's uniform acceleration. The formula for the period of a simple pendulum demonstrates that only the length and gravitational acceleration are significant factors. Additionally, the amplitude does have an effect, but it is minimal for small angles. Air resistance can also play a role, particularly through the exposed area of the pendulum. Overall, the key factors influencing the frequency are length and amplitude, with mass being negligible.
BBboy
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hi, i have a question
"What are the relationships between the frequency of a simple pendulum and it's mass, amplitude abd length?"
im pretty sure only the amplitude affects the pendulum while the mass and length don't right?
 
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Look up (or derive) the formula for the period of a simple pendulum.
 
Im pretty sure only the amplitude affects the pendulum while the mass and length don't right?

Why don't you try it? Tie a ball to a piece of string and have a look.
 
The length does since the arc becomes greater for a same angle. The mass dosen't since for gravity mass dosen't affect the acceleration. Also why not consider the exposed area? That does affect the period since it is related to air resistance.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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