Mechanical Engineering - Theory of machines - Discussion
Discussion Forum : Theory of machines - Section 3 (Q.No. 13)
13.
In under damped vibrating system, if x1 and x2 are the successive values of the amplitude on the same side of the mean position, then the logarithmic decrement is equal to
log
loge
log(x1.x2)
Discussion:
10 comments Page 1 of 1.
Salahuddin said:
3 years ago
Yes, agree. C is the correct answer.
Basavarajappa said:
5 years ago
C is the right answer.
Harshad said:
6 years ago
Option C is the correct answer.
Deepak Sharma said:
7 years ago
The right answer is C. Because the base of log is e in logarithmic decrement. So the LOGe =ln.
Raj said:
8 years ago
lograthmic decreament= 1/n log(x1/x2).
Meet said:
8 years ago
C is the right answer.
Ekam said:
8 years ago
I think C is the right answer.
Vigneshwaran C said:
9 years ago
C is correct,
Logarithmic decrement,
Delta =ln (Xn/Xn + 1).
ln = loge.
Logarithmic decrement,
Delta =ln (Xn/Xn + 1).
ln = loge.
VIKRANT said:
1 decade ago
Yes the correct answer option is C. Because ::::: (x2/x1) = e^(logarithmic decrement).
Jitu said:
1 decade ago
I think so that the answer is C because,
Logarithmic decrement = ln(x1/x2) = loge(x1/x2).
Logarithmic decrement = ln(x1/x2) = loge(x1/x2).
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