Mechanical Engineering - Theory of machines - Discussion
Discussion Forum : Theory of machines - Section 1 (Q.No. 18)
18.
If ω/ωn = 2, where co is the frequency of excitation and ωn is the natural frequency of vibrations, then the transmissibility of vibration will the
Discussion:
5 comments Page 1 of 1.
SHASHANK said:
1 decade ago
TRANSMISSIBILITY = F(t) /F(o).
T = SQURT(1+(2*ZETA * FREQUENCY /NATURAL FREQUENCY)^2 )/SQUART ( (1-(FREQUENCY/NATURAL FREQUENCY)^2 )^2 + ( 2 * ZETA * FREQUENCY/NATURAL FREQUENCY)^2 )
TAKE ZETA = 1.
AND FREQUENCY/NATURAL FREQUENCY = SQURE ROOT 2 (GIVEN).
WILL GET TRANSMISSIBILITY = 1 ANS.
T = SQURT(1+(2*ZETA * FREQUENCY /NATURAL FREQUENCY)^2 )/SQUART ( (1-(FREQUENCY/NATURAL FREQUENCY)^2 )^2 + ( 2 * ZETA * FREQUENCY/NATURAL FREQUENCY)^2 )
TAKE ZETA = 1.
AND FREQUENCY/NATURAL FREQUENCY = SQURE ROOT 2 (GIVEN).
WILL GET TRANSMISSIBILITY = 1 ANS.
Narayan said:
5 months ago
@Suraj Anand.
For any damping factor (zeta) , the transmissibility will remain 1 caz term containing zeta from the numerator and the denominator will get cancelled when you substitute the value of frequency ratio. So that's why Z = 1. Which is arbitrary. You can try with any real number.
For any damping factor (zeta) , the transmissibility will remain 1 caz term containing zeta from the numerator and the denominator will get cancelled when you substitute the value of frequency ratio. So that's why Z = 1. Which is arbitrary. You can try with any real number.
Anshuman said:
5 years ago
Damped frequency /naturalfreuqncy= (1-Ζ^2)^0.5, put values and find Ζ.
Suraj anand said:
8 years ago
Why zeta =1 taken here? Please explain @Shashank.
ARLA said:
8 years ago
e = 1/((w/wn)^2-1). Put e=1. w/wn = 2^(1/2).
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