Show that the period of oscillation on a main helical spring is given by T=2π√{k/m}
Answer:
According to Newton's second law we have: "a=\\dfrac{F}{m}=-\\dfrac{k}{m}x;"
"a=-w^2x;"
Equating the two formulas for acceleration we obtain:
"-w^2x=-\\dfrac{k}{m}x; w^2=\\dfrac{k}{m}; w=\\sqrt{\\dfrac{k}{m}};"
"T=\\dfrac{2\\pi}{\\omega}; T=2\\pi\\sqrt{\\dfrac{m}{k}}."
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