Question

A $200\mathrm{g}$block hangs from a spring with spring constant $10\mathrm{N}/\mathrm{m}$. At the block islocalid="1650020763199" $20\mathrm{cm}$below the equilibrium point and moving upward with a speed oflocalid="1650020788839" $100\mathrm{cm}/\mathrm{s}$. What are the block's

a. Oscillation frequency?

b. Distance from equilibrium when the speed is $50\mathrm{cm}/\mathrm{s}$?

c. Distance from equilibrium at $\mathrm{t}=1.0\mathrm{s}$?

Short answer

a. Oscillation frequency is $1.1\mathrm{Hz}$.

b. Distance is $23\mathrm{cm}$.

c. Distance is$-4.1\mathrm{cm}$.

Step-by-step solution

Calculation for frequency of oscillation (part a)

a.

Given info:

The mass is,

$\mathrm{m}=(200\mathrm{g})\times \left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)$

$=0.2\mathrm{kg}$

The spring constant is$10\mathrm{N}/\mathrm{m}$.

Oscillation frequency is,

$\mathrm{f}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

$\mathrm{f}=\frac{1}{2\times 3.14}\sqrt{\frac{10\mathrm{N}/\mathrm{m}}{0.2\mathrm{kg}}}$

$\mathrm{f}=1.1\mathrm{Hz}$

Calculation for angular frequency and function of time

Angular frequency is,

$\mathrm{\omega }=\sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$

$\mathrm{\omega }=\sqrt{\frac{10\mathrm{N}/\mathrm{m}}{0.2\mathrm{kg}}}$

$\mathrm{\omega }=5\sqrt{2}{\mathrm{s}}^{-1}$

Function of time is,

$\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Acos}(\mathrm{\omega t}+\mathrm{\varphi })$

For time is zero means,

$\mathrm{y}\left(0\right)=\mathrm{Acos}\left(\mathrm{\varphi }\right)$

$=-20\mathrm{cm}$

Calculation of speed

Speed is,

${\mathrm{v}}_{\mathrm{y}}\left(\mathrm{t}\right)=\frac{\mathrm{dy}\left(\mathrm{t}\right)}{\mathrm{dt}}$

$=\frac{\mathrm{d}}{\mathrm{dt}}\left[\mathrm{Acos}\right(\mathrm{\omega t}+\mathrm{\varphi }\left)\right]$

${\mathrm{v}}_{\mathrm{y}}\left(\mathrm{t}\right)=-\mathrm{\omega A}\left[\sin \right(\mathrm{\omega t}+\mathrm{\varphi }\left)\right]$

Where time is zero,

${\mathrm{v}}_{\mathrm{y}}\left(0\right)=-\mathrm{\omega A}\left[\mathrm{sin\varphi }\right)]=100\mathrm{cm}/\mathrm{s}$

From that,

$\frac{{\mathrm{v}}_{\mathrm{y}}\left(0\right)}{\mathrm{y}\left(0\right)}=-\mathrm{\omega tan}\left(\mathrm{\varphi }\right)=\frac{100\mathrm{cm}/\mathrm{s}}{-20\mathrm{cm}}$

$-\mathrm{\omega tan}\left(\mathrm{\varphi }\right)=-5{\mathrm{s}}^{-1}$

$=-2.526\mathrm{rad}$

Calculation for time

Amplitude is ,

$\mathrm{A}=\frac{-20\mathrm{cm}}{\cos (-2.526\mathrm{rad})}$

$=24.5\mathrm{cm}$

Time calculated as,

${\mathrm{v}}_{\mathrm{y}}\left(\mathrm{t}\right)=-\mathrm{\omega Asin}(\mathrm{\omega t}+\mathrm{\varphi })$

$\mathrm{\omega t}+\mathrm{\varphi }={\sin }^{-1}\left(\frac{-50\mathrm{cm}/\mathrm{s}}{\mathrm{\omega A}}\right)$

$\mathrm{t}=\frac{{\sin }^{-1}\left(\frac{-50\mathrm{cm}/\mathrm{s}}{\mathrm{\omega A}}\right)-\mathrm{\varphi }}{\mathrm{\omega }}$

$\mathrm{t}=\frac{{\sin }^{-1}\left[\frac{-50\mathrm{cm}/\mathrm{s}}{\left(5\sqrt{2}{\mathrm{s}}^{-1}\right)(24.5\mathrm{cm})}-(-2.526\mathrm{rad})\right]}{5\sqrt{2}{\mathrm{s}}^{-1}}$

$=0.3158\mathrm{s}$

Calculation of distance (part b)

b.

When speed is $50\mathrm{cm}/\mathrm{s}$,

$\mathrm{x}(0.3158\mathrm{s})=(24.5\mathrm{cm})\cos \left[\left(5\sqrt{2}{\mathrm{s}}^{-1}\right)(0.3158\mathrm{s})-2.526\mathrm{rad}\right]$

$=23\mathrm{cm}$

Calculation for distance (part c)

c.

Where $\mathrm{t}=1.0\mathrm{s}$,

$\mathrm{x}(1.0\mathrm{s})=\mathrm{Acos}(\mathrm{\omega t}+\mathrm{\varphi })$

$=(24.5\mathrm{cm})\cos \left[\left(5\sqrt{2}{\mathrm{s}}^{-1}\right)(1.0\mathrm{s})+2.526\mathrm{rad}\right]$

$=-4.1\mathrm{cm}$