Two pendulums of length I meter and 16 meters start vibrating one behi...
Solution:
Given, length of shorter pendulum = 1 m
Length of longer pendulum = 16 m
Time period of shorter pendulum = T
Let's assume that after time t, the two pendulums are again in the same phase and the thread of the longer pendulum is one behind the shorter one.
Now, we know that the time period of a pendulum is given by:
T = 2π√(l/g)
where l is the length of the pendulum and g is the acceleration due to gravity.
Therefore, time period of shorter pendulum = T = 2π√(1/g)
Time period of longer pendulum = T₁ = 2π√(16/g) = 8T
Let's consider the displacement of the shorter pendulum as x(t) and that of the longer pendulum as y(t). At t=0, both pendulums are in the same phase and the amplitude of both the pendulums is zero.
Now, we know that the displacement of a simple pendulum is given by:
x(t) = Asin(ωt + φ)
where A is the amplitude, ω is the angular frequency, and φ is the phase constant.
At t=0, both the pendulums are in the same phase and the amplitude of both the pendulums is zero. Therefore, the equation for the displacement of both the pendulums at t=0 is:
x(0) = y(0) = 0
Now, let's assume that after time t, the two pendulums are again in the same phase and the thread of the longer pendulum is one behind the shorter one. Therefore, the displacement of the shorter pendulum at time t is given by:
x(t) = Asin(ωt)
where ω = 2π/T
The displacement of the longer pendulum at time t is given by:
y(t) = Asin(2πt/T₁ + φ)
Now, we need to find the minimum time t after which the thread of the longer pendulum is one behind the shorter one. This means that the displacement of the longer pendulum is π/2 radians behind the displacement of the shorter pendulum.
Therefore, we have:
y(t) = x(t - δ)
where δ = π/2ω
Substituting the values of x(t) and y(t), we get:
Asin(2πt/T₁ + φ) = Asin(ω(t - δ))
Solving this equation, we get:
2πt/T₁ + φ = ω(t - δ) + nπ
where n is an integer
Substituting the values of T₁, ω, and δ, we get:
2πt/8T + φ = πt/T - π/2 + nπ
Simplifying this equation, we get:
t = (2n + 1)T/5
Therefore, the minimum time after which the two threads of the pendulums will be one behind the other is (2n + 1)T/5, where n is an integer.
Since we need to find the minimum time, we take n=0. Therefore, the answer is:
t = T/5 + T
t = 2T/5
Hence, the correct answer is option (a
Two pendulums of length I meter and 16 meters start vibrating one behi...
Shouldnt the answer be d
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