* Frequency of a
as
a tuning fork
is
300 Hz and its Q factor is 5×10^4
. Find the relaxation time
Also find the
time
after which it's
energy
becomes (1/10)lits of the it's initial undapmed value.
Answers
Answered by
10
The time constant is 26.5
Explanation:
We are gievn that:
- Frequency of tunning fork = 300 Hz
- The value of Q factor = 5×10^4
- Q = 2π = maximum energy stored per cycle/ energy dissipated per cycle
- Q = ωL / R
- Q = 2πf L / R => L / R = Q / 2πf
- L / R time constant = 5 x 10^4 / 2 x 3.14 x 300
- L / R time constant = 5x 10^4 / 1884
- L / R time constant = 26.5
- Thus the time constant is 26.5
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