Physics 9702 · Paper 1

Springs in Series & Parallel

Combining springs changes the effective stiffness $k_{\text{eff}}$ . The way you connect them determines whether the set is softer or stiffer.
Series (end to end): softer, so for two identical springs $k_{\text{eff}}=k/2$ .
Parallel (side by side): stiffer, so for two identical springs $k_{\text{eff}}=2k$ .

Series: end to end, both springs feel the full load, so the extensions add and k_eff = k/2. Parallel: side by side, each spring carries half the load, so k_eff = 2k.

Where the two rules come from

k_{\text{eff}}=\tfrac{k}{2}\ \text{(series)} \qquad k_{\text{eff}}=2k\ \text{(parallel)}

Symbols

$k$ = spring constant of one spring (N m⁻¹)
$k_{\text{eff}}$ = effective combined spring constant (N m⁻¹)

Series. The same force $F$ acts in each spring, so each stretches by $F/k$ , and the extensions add: $x=2(F/k)$ . Then $k_{\text{eff}}=F/x=k/2$ . In general $\dfrac{1}{k_{\text{eff}}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$ . Series springs are softer.
Parallel. Both springs share the same extension $x$ , so each pulls with $kx$ and together they support $2kx$ . Then $k_{\text{eff}}=(2kx)/x=2k$ . In general $k_{\text{eff}}=k_1+k_2$ . Parallel springs are stiffer.

To check your answer: the set that stretches more under the same load is the softer, series set ( $k/2$ ).

Worked examples: parallel and series

Worked example

Two identical springs combined

Two identical springs each have $k=200$ N m^-1.

In parallel the stiffness doubles: $k_{\text{eff}}=2k=400\text{ N m}^{-1}$ .
In series it halves: $k_{\text{eff}}=k/2=100\text{ N m}^{-1}$ .

Answer

\text{parallel }400\text{ N m}^{-1}\quad\text{series }100\text{ N m}^{-1}

Worked example

Adding a second spring in parallel

A cable holds a load with tension 110 N. A spring of constant 250 N m^-1 is first used alone; then a second identical spring is added in parallel.

Parallel doubles the stiffness: $k_{\text{eff}}=2\times250=500\text{ N m}^{-1}$ .
$x=F/k_{\text{eff}}=110/500=0.22\text{ m}$ .

Answer

x=0.22\text{ m}

(9702/22/O/N/25 Q3(c))

Worked example

Two different springs in series

Springs of 6.0 N cm^-1 and 4.0 N cm^-1 are joined in series and stretched by a load of 80 N. Find the total extension.

Series adds the reciprocals: $\dfrac{1}{k}=\dfrac{1}{6.0}+\dfrac{1}{4.0}$ .
$k=2.4\text{ N cm}^{-1}$ .
$x=F/k=80/2.4=33\text{ cm}$ .

Answer

x=33\text{ cm (D)}

(9702/12/F/M/24 Q23)

Common mistake

Series and parallel are easy to confuse. Series springs are softer (

k/2

); parallel springs are stiffer (

2k

). If you add the reciprocals you are in series; if you add the constants you are in parallel.

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Springs in Series & Parallel

Combining springs changes the effective stiffness $k_{\text{eff}}$ . The way you connect them determines whether the set is softer or stiffer.
Series (end to end): softer, so for two identical springs $k_{\text{eff}}=k/2$ .
Parallel (side by side): stiffer, so for two identical springs $k_{\text{eff}}=2k$ .

Series: end to end, both springs feel the full load, so the extensions add and k_eff = k/2. Parallel: side by side, each spring carries half the load, so k_eff = 2k.

Where the two rules come from

k_{\text{eff}}=\tfrac{k}{2}\ \text{(series)} \qquad k_{\text{eff}}=2k\ \text{(parallel)}

Symbols

$k$ = spring constant of one spring (N m⁻¹)
$k_{\text{eff}}$ = effective combined spring constant (N m⁻¹)

Series. The same force $F$ acts in each spring, so each stretches by $F/k$ , and the extensions add: $x=2(F/k)$ . Then $k_{\text{eff}}=F/x=k/2$ . In general $\dfrac{1}{k_{\text{eff}}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$ . Series springs are softer.
Parallel. Both springs share the same extension $x$ , so each pulls with $kx$ and together they support $2kx$ . Then $k_{\text{eff}}=(2kx)/x=2k$ . In general $k_{\text{eff}}=k_1+k_2$ . Parallel springs are stiffer.

To check your answer: the set that stretches more under the same load is the softer, series set ( $k/2$ ).

Worked examples: parallel and series

Worked example

Two identical springs combined

Two identical springs each have $k=200$ N m^-1.

In parallel the stiffness doubles: $k_{\text{eff}}=2k=400\text{ N m}^{-1}$ .
In series it halves: $k_{\text{eff}}=k/2=100\text{ N m}^{-1}$ .

Answer

\text{parallel }400\text{ N m}^{-1}\quad\text{series }100\text{ N m}^{-1}

Worked example

Adding a second spring in parallel

A cable holds a load with tension 110 N. A spring of constant 250 N m^-1 is first used alone; then a second identical spring is added in parallel.

Parallel doubles the stiffness: $k_{\text{eff}}=2\times250=500\text{ N m}^{-1}$ .
$x=F/k_{\text{eff}}=110/500=0.22\text{ m}$ .

Answer

x=0.22\text{ m}

(9702/22/O/N/25 Q3(c))

Worked example

Two different springs in series

Springs of 6.0 N cm^-1 and 4.0 N cm^-1 are joined in series and stretched by a load of 80 N. Find the total extension.

Series adds the reciprocals: $\dfrac{1}{k}=\dfrac{1}{6.0}+\dfrac{1}{4.0}$ .
$k=2.4\text{ N cm}^{-1}$ .
$x=F/k=80/2.4=33\text{ cm}$ .

Answer

x=33\text{ cm (D)}

(9702/12/F/M/24 Q23)

Common mistake

Series and parallel are easy to confuse. Series springs are softer (

k/2

); parallel springs are stiffer (

2k

). If you add the reciprocals you are in series; if you add the constants you are in parallel.