Math, asked by patnaiksagar98, 3 months ago

Two spring elements K1 and K2 are arranged in series and the combination is in parallel with the third spring K3. What is the equivalent spring stiffness of the system?

Answers

Answered by ramsmedicine
2

Answer:

Two springs of spring constant k  

1

​  

 and k  

2

​  

 are joined in series.

The effective spring constant of the combination is -

We know using hooke's law

k  

1

​  

 

f  

1

​  

 

​  

=x  

1

​  

 

k  

2

​  

 

f  

2

​  

 

​  

=x  

2

​  

 

X=  

k  

1

​  

 

f  

1

​  

 

​  

+  

k  

2

​  

 

f  

2

​  

 

​  

 

X=  

k  

1

​  

 

F

​  

+  

k  

2

​  

 

F

​  

 

X=F(  

k  

1

​  

 

1

​  

+  

k  

2

​  

 

1

​  

)

X=F[  

k  

1

​  

k  

2

​  

 

(k  

1

​  

+k  

2

​  

)

​  

]

F=  

(k  

1

​  

+k  

2

​  

)

k  

1

​  

k  

2

​  

 

​  

X

Therefore,k=  

(k  

1

​  

+k  

2

​  

)

k  

1

​  

k  

2

​  

 

Step-by-step explanation:

Answered by bhartivb200
1

Answer:

Rq=2/3

Step-by-step explanation:

thanks... i think I'm right if wrong then sorry!

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