Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
As close as I can read it, it appears to be
1/12 gram/second
(0.08333... gm/sec)
The intense heat in the earth's core that causes molten rock in the mantle layer to move.
Answer:
13.33m/s
Explanation:
Given data
m1= 2000kg
u1= 20m/s
m2= 1500kg
u2= 0m/s
v1= 10m/s
Required
The speed of the sticks
We know that from the expression for the conservation of momentum
m1u1+m2u2= m1v1+m2v2
2000*20+1500*0=2000*10+1500*v2
40000=20000+1500v2
collect like terms
40000-20000= 1500v2
20000= 1500v2
v2= 20000/1500
v2= 13.33 m/s
Hence the velocity of the sticks is 13.33m/s
