[ 3 Marks Questions ]

Question 13 Marks

Derive the expression for resultant spring constant when two springs having constant $k_1$ and $k_2$ are connected in parallel.

Answer

$k_1$ and $k_2$ attached to a mass $m$ as shown in figure. The results can be generalized to any number of springs in parallel.

Let the force F be applied towards right as shown in figure. In this case, both the springs elongate or compress by the same amount of displacement. Therefore, net force for the displacement of mass m is
$F=-k_p X \ldots(1)$
where $k_p$ is called effective spring constant.
Let the first spring be elongated by a displacement $x$ due to force $F_1$ and second spring be elongated by the same displacement x due to force $F _2$, then the net force
$F=-k_1 x-k_2 x$
Equating equations (2) and (1), we get
$k_p=k_1+k_2 \ldots . .(2)$
Generalizing, for $n$ springs connected in parallel,
$k_P=\sum_{i=1}^n k_i$
If all spring constants are identical i.e., $k_1=k_2=\ldots=k_n=k$ then
$k_P=n k$
This implies that the effective spring constant increases by a factor $n$. Hence, for the springs in parallel connection, the effective spring constant is greater than individual spring constant.

View full question & answer→

Question 23 Marks

Derive the expression for resultant spring constant when two springs having constant $k _1$ and $k _2$ are connected in series.

Answer

Let $x_1$ and $x_2$ be the elongation of springs from their equilibrium position (un-stretched position) due to applied force $F$. Then, the net displacement of the mass point is

Therefore, substituting equation (3) in equation (2), the effective spring constant can be calculated as
$\begin{aligned}-\frac{ F }{k_1}-\frac{ F }{k_2} & =-\frac{ F }{k_s} \quad \text { } \\ \frac{1}{k_s} & =\frac{1}{k_1}+\frac{1}{k_2} \text { (or) } k_s=\frac{k_1 k_2}{k_1+k_2} Nm ^{-1}\end{aligned}$ ..(4)
Suppose we have n springs connected in series, the effective spring constant in series is
$
\frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\ldots+\frac{1}{k_n}=\sum_{i=1}^n \frac{1}{k_i} ...(5)
$
If all spring constants are identical i.e., $k_1=k_2=\ldots=k_n=k$ then
$
\frac{1}{k_s}=\frac{n}{k} \Rightarrow k_s=\frac{k}{n} ...(6)
$
This means that the effective spring constant reduces by the factor $n$. Hence, for springs in series connection, the effective spring constant is lesser than the individual spring constants. From equation (3), we have,
$
k_1 x_1=k_2 x_2>
$
Then the ratio of compressed distance or elongated distance $x _1$ and $x _2$ is
$
\frac{x_2}{x_1}=\frac{k_1}{k_2} ...(7)
$
The elasticity potential energy stored in first and second springs are $V _1=\frac{1}{2} k_1 x_1^2$ and $V _2=\frac{1}{2} k_2 x_2^2$ respectively. Then their ratio is
$
\frac{ V _1}{ V _2}=\frac{\frac{1}{2} k_1 x_1^2}{\frac{1}{2} k_2 x_2^2}=\frac{k_1}{k_2}\left(\frac{x_1}{x_2}\right)^2=\frac{k_2}{k_1}..(8)
$

View full question & answer→

Physics [ 3 Marks Questions ]

Test yourself on this topic