the amplitude of an oscillator decreases to 36.8% of its initial value in 10.0 s. what is the value of the time constant
1 answer:
Answer:
τ = 5 s
Explanation:
When a vibrating body is damped. Its amplitude starts to decrease. This decrement is exponential. And it is given as follows:
where,
τ = Time Constant = ?
X = Instantaneous value of amplitude
X₀ = Initial Value of amplitude
t = time interval = 10 s
The ratio of decrement is given as:
therefore, using these values, we get:
Taking natural log (ln) on both sides, we get:
<u>τ = 5 s</u>
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Answer:
t_{out} = t_{in}, t_{out} =
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D /
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D /
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D /
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =