Question

A spring has a 12 cm length. When a 200-g mass is hung from the spring, it extends to 27 cm. The hanging mass were pulled downward a further 5 cm. What is the time-dependent function of the position, in centimeters, of the mass, assuming that the phase angle LaTeX: \phi=0ϕ=0?

LaTeX: x(t)=15\cos{(8.08\textrm{ }t)}x(t)=15cos⁡(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.

LaTeX: x(t)=5\cos{(8.08\textrm{ }t)}x(t)=5cos⁡(8.08 t), where the coefficient inside the trigonometric function has units of rad/s.

LaTeX: x(t)=5\cos{(1.29\textrm{ }t)}x(t)=5cos⁡(1.29 t)x(t)=5cos⁡(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.

LaTeX: x(t)=15\cos{(1.29\textrm{ }t)}x(t)=15cos⁡(1.29 t)x(t)=15cos⁡(1.29 t), where the coefficient inside the trigonometric function has units of rad/s.

Answer 1

Answer:

The equation of the time-dependent function of the position is $x(t)=5\cos(8.08t)$

(b) is correct option.

Explanation:

Given that,

Length = 12 cm

Mass = 200 g

Extend distance = 27 cm

Distance = 5 cm

Phase angle =0°

We need to calculate the spring constant

Using formula of restoring force

$F=kx$

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{200\times10^{-3}\times9.8}{(27-12)\times10^{2}}$

$k=13.06\ N/m$

We need to calculate the time period

Using formula of time period

$T=2\pi\sqrt{\dfrac{m}{k}}$

Put the value into the formula

$T=2\pi\sqrt{\dfrac{0.2}{13.6}}$

$T=0.777\ sec$

At t = 0, the maximum displacement was 5 cm

So, The equation of the time-dependent function of the position

$x(t)=A\cos(\omega t)$

Put the value into the formula

$x(t)=5\cos(2\pi\times f\times t)$

$x(t)=5\cos(2\pi\times\dfrac{1}{T}\times t)$

$x(t)=5\cos(2\pi\times\dfrac{1}{0.777}\times t)$

$x(t)=5\cos(8.08t)$

Hence, The equation of the time-dependent function of the position is $x(t)=5\cos(8.08t)$