A thin metal disk of mass m=2.00 x 10^-3 kg and radius R=2.20cm is attached at its center to a long fiber. When the disk is turn
ed from the relaxed state through a small angle theta, the torque exerted by the fiber on the disk is proportional to theta. Find an expression for the torsional constant k in terms of the moment of inertia I of the disk and the angular frequency w of small, free oscillations. The disk, when twisted and released, oscillates with a period T of 1.00 s. Find the torsional constant k of the fiber.
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Answer:
Explanation:
Given that:
x(t) = 10 sin(10t) . sin (15t)
the objective is to find the power and the rms value of the following signal square.
Recall that:
sin (A + B) + sin(A - B) = 2 sin A.cos B
x(t) = 10 sin(15t) . cos (10t)
x(t) = 5(2 sin (15t). cos (10t))
x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)
x(t) = 5sin(25 t) + 5 sin (5t)
From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:
For the number of sinosoidial signals;
Power can be expressed as:
As such,
For x(t), Power
For the number of sinosoidial signals;
For x(t), the RMS value is as follows:
Answer:
attached below
Explanation:
