Physics

Problem

19. A mass

m

is attached to a vertical spring stretching it distance

d

. Then, the mass is set oscillating on a spring with an amplitude of A, the period of oscillation is proportional to (A)

\sqrt{\frac{d}{g}}

(B)

\sqrt{\frac{g}{d}}

(C)

\sqrt{\frac{d}{m g}}

(D)

\sqrt{\frac{m^{2} g}{d}}

(A)

\sqrt{\frac{d}{g}}

(B)

\sqrt{\frac{g}{d}}

(C)

\sqrt{\frac{d}{m g}}

(D)

\sqrt{\frac{m^{2} g}{d}}

Intuitive explanation

The key insight is that the equilibrium stretch $d$ encodes information about the spring constant. When the mass hangs at rest, $mg = kd$ , so $k = mg/d$ . The period of a spring oscillator depends only on $k$ and $m$ , not on amplitude—so use $T = 2\pi\sqrt{m/k}$ and substitute your expression for $k$ in terms of the given quantities $d$ , $m$ , and $g$ . You'll find the period is proportional to $\sqrt{d/g}$ .

Intuitive explanation

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