Two springs A and B ( $k_{A}$ = 2 $k_{B}$ ) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is

Q. Two springs

A

and

B

are stretched by applying forces of equal magnitudes. If the energy stored in

A

is E, then the energy stored in B will be-
(Given relation between spring constants of

A

and

B

(k_{A} = 2 k_{B})

)

Q. Two springs A and B (

k_{A} = 2 k_{B}

) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, then in B is (assume equilibrium):

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Two springs have force constants kA and kB such that kB=2kA. The four ends of the springs are stretched by the same force. If energy stored in spring A is E, then energy stored in spring B is

The correct option is C E2F=kx⇒x=FkNow, U=12kx2=12k(Fk)2or Uα1kkB is double. Therefore, UB will be half.

Two springs have force constants $k_{A}$ and $k_{B}$ such that $k_{B} = 2 k_{A}$ . The four ends of the springs are stretched by the same force. If energy stored in spring A is $E$ , then energy stored in spring B is

The correct option is C $\frac{E}{2}$
$F = k x \Rightarrow x = \frac{F}{k}$

Now, $U = \frac{1}{2} k x^{2}$

$= \frac{1}{2} k (\frac{F}{k})^{2}$

or $U α \frac{1}{k}$

$k_{B}$ is double. Therefore, $U_{B}$ will be half.