dua pegas masing masing ditarik dengan gaya yang sama besar 4N .Bila konstan pegas pertama 1/4 kali konstanta pegas kedua ,maka perbandingan pertambahan panjang

F1=F2= 4N; k1=1/4k2
F= kΔx
maka Δx1: Δx2= F1/k1 : F2/ k2
= 4/1/4 k2 : 4/1 k2
= 16 k2 :4k2
→4:1

[tex]\displaystyle F_1=F_2=4\,\text{N}\\k_1=\frac14k_2\\\\\Delta x_1:\Delta x_2=?\\\\\Delta x_1:\Delta x_2=\frac{F_1}{k_1}:\frac{F_2}{k_2}\\\Delta x_1:\Delta x_2=\frac{4}{\frac14k_2}:\frac{4}{k_2}\\\Delta x_1:\Delta x_2=\frac{1}{\frac14}:\frac{1}{1}\\\boxed{\boxed{\Delta x_1:\Delta x_2=4:1}}[/tex]