Mechanical Engineering - Theory of machines - Discussion
Discussion Forum : Theory of machines - Section 1 (Q.No. 49)
49.
A mass of 1 kg is attached to the end of a spring with a stiffness of 0.7 N/mm. The critical damping coefficient of this system is
Discussion:
7 comments Page 1 of 1.
Kapil Somani said:
1 decade ago
Critical damping coefficient = 2*Sqrt(KM).
= 2*Sqrt(700*1).
= 52.92 N-s/m.
= 2*Sqrt(700*1).
= 52.92 N-s/m.
Samsad ali said:
10 years ago
Please explain it.
Zeyaul said:
10 years ago
Answer is wrong.
Because k = 0.7 n/m not 0.7 kn/m.
Because k = 0.7 n/m not 0.7 kn/m.
Suresh said:
9 years ago
Answer is correct because k is in N/mm. If we convert that into N/m it should be 700N/m.
Chahat said:
8 years ago
It is 2 * √(k*m).
K = 0.7N/mm = 700 N/m.
K = 0.7N/mm = 700 N/m.
Tesfay said:
7 years ago
K=0.7N/mm=700N/m M=1kg Coefficient=2.Sqrt(KM). =2.Sqrt (700*1). = 52.92 N-s/m
Viresh c said:
6 years ago
(critical damping coefficient)Cc=2√km. (formula).
Given
K=0.7N/mm=700N/m
m =1kg.
So, Cc=2√km.
=2√700*1
Cc=52.92N-s/m.
Given
K=0.7N/mm=700N/m
m =1kg.
So, Cc=2√km.
=2√700*1
Cc=52.92N-s/m.
(1)
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