Question

the frequency of a tuning fork is 300 hz and its q favtor is 5×10^4. find the relaxation time
also find the line after which the energy becomes 1/10 of its initial damped value
(E=Eoe^-2kt,w=2πf,t=61.12s)​

Answer 1

$\huge\red{Question}$

the frequency of a tuning fork is 300 hz and its q favtor is 5×10^4. find the relaxation time

also find the line after which the energy becomes 1/10 of its initial damped value

(E=Eoe^-2kt,w=2πf,t=61.12s)​

$\huge\purple{answer}$

The time constant is 26.5

$\huge{e}\red{x}\blue{p}{l}\green{a}{n}\pink{a}{t}\purple{i}{o}\orange{n}$

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

Answer 2

$\huge \red{question}$

the frequency of a tuning fork is 300 hz and its q favtor is 5×10^4. find the relaxation time also find the line after which the energy becomes 1/10 of its initial damped value (E=Eoe^-2kt,w=2πf,t=61.12s)

$\huge\purple{answer}$

The time constant is 26.5

$\huge{e}\red{x}\blue{p}{l}\green{a}{n}\pink{a}{t}\purple{i}{o}\orange{n}$

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