Question

* 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.​

Answer 1

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