Question

Astronauts in space cannot weigh themselves by standing on a bathroom scale. Instead, they determine their mass by oscillating on a large spring. Suppose an astronaut attaches one end of a large spring to her belt and the other end to a hook on the wall of the space capsule. A fellow astronaut then pulls her away from the wall and releases her. The spring's length as a function of time is shown in FIGURE$\mathrm{P}15.43$.

a. What is her mass if the spring constant is$240\mathrm{N}/\mathrm{m}$$?$

b. What is her speed when the spring's length is$1.2\mathrm{m}$$?$

Short answer

a. The mass spring constant is$55\mathrm{kg}$.

b. The speed is$0.73\frac{\mathrm{m}}{\mathrm{s}}$.

Step-by-step solution

Calculation for mass (part a)

a.

In simple harmonic motion, the period of an oscillator is given by

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

From figure,

$\mathrm{T}=3\mathrm{s}$

So,

${\mathrm{T}}^2=4{\mathrm{\pi }}^2\left(\frac{\mathrm{m}}{\mathrm{k}}\right)$

$\mathrm{m}=\frac{{\mathrm{kT}}^2}{4{\mathrm{\pi }}^2}$

$\mathrm{m}=\frac{(240\mathrm{N}/\mathrm{m})(3\mathrm{s}{)}^2}{4{\left(3.14\right)}^2}$

$\mathrm{m}=55\mathrm{Kg}$

Calculation for area and frequency (part b) 

b.

From figure,

Area is,

$\mathrm{A}=\frac{1.4\mathrm{m}-0.6\mathrm{m}}{2}$

$=0.4\mathrm{m}$

Angular frequency,

$\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}$

$\mathrm{\omega }=\frac{2\mathrm{\pi }}{3}{\mathrm{s}}^{-1}$

Calculation for time part (b) solution

b.

Maximum displacement is,

$\mathrm{x}\left(\mathrm{t}\right)=1.0\mathrm{m}+\mathrm{Acos}\left(\mathrm{\omega t}\right)$

Where $\mathrm{X}\left(\mathrm{t}\right)=1.2\mathrm{m}$

$1.2\mathrm{m}=1.0\mathrm{m}+(0.4\mathrm{m})\cos \left[\left(\frac{2\mathrm{\pi }}{3}{\mathrm{s}}^{-1}\right)\mathrm{t}\right]$

$0.2\mathrm{m}=(0.4\mathrm{m})\cos \left[\left(\frac{2\mathrm{\pi }}{3}{\mathrm{s}}^{-1}\right)\mathrm{t}\right]$

$0.5=\cos \left[\left(\frac{2\mathrm{\pi }}{3}{\mathrm{s}}^{-1}\right)\mathrm{t}\right]$

${\cos }^{-1}(0.5)=\left(\frac{2\mathrm{\pi }}{3}{\mathrm{s}}^{-1}\right)\mathrm{t}$

role="math" localid="1650017451513" $\mathrm{t}=\frac{{\cos }^{-1}(0.5)}{2\mathrm{\pi }/3{\mathrm{s}}^{-1}}$

$=0.5\mathrm{s}$

Calculation for speed part (b) solution

b.

Speed is,

${\mathrm{v}}_{\mathrm{x}}=\frac{\mathrm{dx}\left(\mathrm{t}\right)}{\mathrm{dt}}$

$=\frac{\mathrm{d}}{\mathrm{dt}}[1.0\mathrm{m}+\mathrm{Acos}(\mathrm{\omega t}\left)\right]$

$=-\mathrm{\omega Asin}\left(\mathrm{\omega t}\right)$

Speed has magnitude velocity of ,

$=\left|-\mathrm{\omega Asin}\left(\mathrm{\omega t}\right)\right|$

So,

$\mathrm{v}=\left|-\left(\frac{2\mathrm{\pi }}{3}{\mathrm{s}}^{-1}\right)(0.4\mathrm{m})\sin \left[\left(\frac{2\mathrm{\pi }}{3}{\mathrm{s}}^{-1}\right)(0.5\mathrm{s})\right]\right|$

$\mathrm{v}=0.73\frac{\mathrm{m}}{\mathrm{s}}$