A block attached to a spring is made oscillate with initial amplitude of 8.0 cm after 2.2 minutes, the amplitude decrease to 5.0 cm calculate
A0=8cm=0.08mA_0 = 8cm = 0.08mA0=8cm=0.08m
A1=5cm=0.05mA_1 = 5cm = 0.05mA1=5cm=0.05m
t1=2.2min=132st_1 = 2.2min = 132st1=2.2min=132s
A2=2cm=0.02mA_2 = 2cm = 0.02mA2=2cm=0.02m
A1=A0e−γt1A _1= A_0e^{-\gamma t_1}A1=A0e−γt1
γ=−1t1∗lnA1A0=−1132∗ln0.050.08=3.56∗10−3\gamma=-\frac{1}{t_1}*\ln\frac{A_1}{A_0}= -\frac{1}{132}*\ln \frac{0.05}{0.08}=3.56*10^{-3}γ=−t11∗lnA0A1=−1321∗ln0.080.05=3.56∗10−3
A2=A0e−γt2A _2= A_0e^{-\gamma t_2}A2=A0e−γt2
t2=−1γ∗lnA2A0=−13.56∗10−3∗ln0.020.08≈389st_2=-\frac{1}{\gamma}*\ln\frac{A_2}{A_0}= -\frac{1}{3.56*10^{-3}}*\ln \frac{0.02}{0.08}\approx389st2=−γ1∗lnA0A2=−3.56∗10−31∗ln0.080.02≈389s
Answer:\text{Answer:}Answer:
(i) t2=389s\text{(i) }t_2=389s(i) t2=389s
(ii) γ=3.56∗10−3\text{(ii) }\gamma=3.56*10^{-3}(ii) γ=3.56∗10−3
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