
The amplitude of a damped oscillator decreases to 0.9 times its original magnitude is 5 seconds. In another 10 seconds it will decrease to ‘a’ times its original magnitude where $\alpha $ equals
A. 0.7
B. 0.81
C. 0.729
D. 0.6
Answer
161.4k+ views
Hint: In case of damped oscillations, first try to find the relation between the amplitude at time t and the initial amplitude. Then put the values of both the amplitude and the time t is given and finally using that time value find the value of amplitude after 10 sec time.
Formula used
$A = {A_0}{e^{ - \alpha t}}$
Where, A is the amplitude at a given time t.
And ${A_0}$is the initial amplitude.
Complete answer:
For case 1: t = 5 seconds
Given the amplitude at time t = 5 seconds becomes 0.9 times of its initial amplitude.
$A = 0.9{A_0}$
Putting this value in the formula, we get;
$0.9{A_0} = {A_0}{e^{ - 5\alpha }}$
After solving, we get: ${e^{ - 5\alpha }} = 0.9$ (equation 1)
For case 2: t= 10 seconds
$a{A_0} = {A_0}{e^{ - 10\alpha }}$
After solving the above equation, we get;
${e^{ - 10\alpha }} = a$ (equation 2)
Solving equation 1 and 2, we get;
${e^{ - 10\alpha }} = a = {({e^{ - 5\alpha }})^2}$
$a = {(0.9)^2} = 0.81$
${e^{ - 10\alpha }} = 0.81$
After solving, we get;
$\alpha = 0.729$
Hence, the correct answer is Option C.
Note:Be careful about the change in amplitude according to the time given and how many times it becomes of the initial amplitude or the maximum amplitude. Use the same formula for both the case at t = 5 seconds and t = 10 seconds. Finally solving both cases together gets the value of a.
Formula used
$A = {A_0}{e^{ - \alpha t}}$
Where, A is the amplitude at a given time t.
And ${A_0}$is the initial amplitude.
Complete answer:
For case 1: t = 5 seconds
Given the amplitude at time t = 5 seconds becomes 0.9 times of its initial amplitude.
$A = 0.9{A_0}$
Putting this value in the formula, we get;
$0.9{A_0} = {A_0}{e^{ - 5\alpha }}$
After solving, we get: ${e^{ - 5\alpha }} = 0.9$ (equation 1)
For case 2: t= 10 seconds
$a{A_0} = {A_0}{e^{ - 10\alpha }}$
After solving the above equation, we get;
${e^{ - 10\alpha }} = a$ (equation 2)
Solving equation 1 and 2, we get;
${e^{ - 10\alpha }} = a = {({e^{ - 5\alpha }})^2}$
$a = {(0.9)^2} = 0.81$
${e^{ - 10\alpha }} = 0.81$
After solving, we get;
$\alpha = 0.729$
Hence, the correct answer is Option C.
Note:Be careful about the change in amplitude according to the time given and how many times it becomes of the initial amplitude or the maximum amplitude. Use the same formula for both the case at t = 5 seconds and t = 10 seconds. Finally solving both cases together gets the value of a.
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