Two springs having spring constants k_1 and k | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For parallel connection,the equivalent spring constant is $k_p = k_1 + k_2 = 9 \dots (1)$For series connection,the equivalent spring constant is $

Vedclass · Accepted Answer

$6 	ext{ unit}, 3 	ext{ unit}$

Vedclass · Answer

(A) For parallel connection,the equivalent spring constant is $k_p = k_1 + k_2 = 9 \dots (1)$For series connection,the equivalent spring constant is $k_s = \frac{k_1 k_2}{k_1 + k_2} = 2 \dots (2)$Substituting equation $(1)$ into equation $(2)$:$\frac{k_1 k_2}{9} = 2 \implies k_1 k_2 = 18 \dots (3)$From $(1)$,$k_2 = 9 - k_1$. Substituting this into $(3)$:$k_1(9 - k_1) = 18$$9k_1 - k_1^2 = 18$$k_1^2 - 9k_1 + 18 = 0$Factoring the quadratic equation:$(k_1 - 6)(k_1 - 3) = 0$Thus,$k_1 = 6 	ext{ unit}$ or $k_1 = 3 	ext{ unit}$.If $k_1 = 6$,then $k_2 = 3$. If $k_1 = 3$,then $k_2 = 6$.Therefore,the values are $6 	ext{ unit}$ and $3 	ext{ unit}$.

Two springs having spring constants $k_1$ and $k_2$ are connected in series,their resultant spring constant is $2 \text{ unit}$. If they are connected in parallel,their resultant spring constant is $9 \text{ unit}$. Find the values of $k_1$ and $k_2$.

Similar Questions

$A$ block is attached to a spring as shown and very gradually lowered so that finally the spring expands by $d$. If the same block is attached to the spring and released suddenly,then the maximum expansion in the spring will be:

An elastic spring under a tension of $3 \,N$ has a length $a$. Its length is $b$ under a tension of $2 \,N$. For its length $(3a - 2b)$, the value of tension will be . . . . . . $N$.

The block $A$ is pushed towards the wall by a distance $x$ and released. The normal reaction $N$ by the vertical wall on the block $B$ versus the compression $x$ in the spring is given by:

The tension in the spring is

$A$ body of mass $m$ is suspended from an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards is

Vedclass Test Series

Exam Paper Generator

Online Exam Module