+91-87964 74404
FlagBACK
Question icon
IIT JEE Entrance Exam

spring stretches from by 1.8 cm and 2.8 kg block is suspended from its end the mass that has to be attached to the spring so that its frequency of the vibration is three Hz is

Profile image of sudhanva gv
7 Years agoGrade
Answers icon

1 Answer

Profile image of TARUN SHARMA
7 Years ago
WHEN mass 2.8 kg attached then the system is in equalibrium with the spring
  so ; kx = mg
k=mg/x
 
and as we know frequency= omega/2pi
 
omega^2=k/m1
 
m1 total mass
by equating values in equation you will get your answer
 

Other Related Questions on iit jee entrance exam

question mark

Let S={x:x belongs to R(√3+√2)^(x^2-4) - (√3-√2)^(x^2-4)=10}. Then n(S) is equal to-

iit jee entrance exam
1 Answer Available

Last Activity: 10 Months ago

question mark
My rank is 29k, General, should I opt for any state college cse branch or some Nits in which in which I am getting ECE , chemical, mechanics etc....
 
iit jee entrance exam
1 Answer Available

Last Activity: 3 Years ago

question mark
How is Ranquer Eduzone,comparing, Akash,Fitjee,Resonance,Naryana
iit jee entrance exam
1 Answer Available

Last Activity: 4 Years ago

question mark
Can class 9th students study syllabus class  of 11th 
iit jee entrance exam
3 Answers Available

Last Activity: 4 Years ago

question mark
Sum of series 2[n/2] ! [n/2]!/ n! [2Co-2C2+3C2....+[-1]^n[n+1]2Cn] where n is an even positive integer. Kindly provide solution of this question and how to generalise this. How do we know that which expansion will give this?
iit jee entrance exam
1 Answer Available

Last Activity: 4 Years ago