Question

A $200\mathrm{g}$mass attached to a horizontal spring oscillates at a frequency of $2.0\mathrm{Hz}$. At $t=0\mathrm{s}$, the mass is at $x=5.0\mathrm{cm}$and has $v_x=-30\mathrm{cm}/\mathrm{s}$. Determine:

$a.$The period.

$b.$The angular frequency.

$c.$The amplitude.

$d.$The phase constant.

$e.$The maximum speed.

$f.$The maximum acceleration.

$g.$The total energy.

$h.$The position at $t=0.40\mathrm{s}$.

Short answer

$a)$The period is $T=0.5s$

$b)$The angular frequency is $\omega =4\mathrm{\pi }$

$c)$The amplitude is$A=5.5cm$

$d)$The phase constant is${\varphi }_0=0.137\mathrm{\pi }$

$e)$The maximum speed is$v_{max}=69cm$

$f)$The maximum acceleration is$a_{\max }=869\mathrm{cm}·{\mathrm{s}}^{-2}$

$g)$The total energy is$E=0.05J$

$h)$The position at$t=0.40s$is$V\left(0.4\right)=50.7cm/s$

Step-by-step solution

Given.

Given

• $m=0.2\mathrm{kg}$
• $f=2\mathrm{Hz}$
• $x\left(0\right)=5\mathrm{cm}$
• $v_x=-30\mathrm{cm}$

Required .

$a)T$

$b)\omega$

$c)A$

$d){\varphi }_0$

$e)v_{max}$

$f)a_{max}$

$g)E$

$h)v\left(0.4\right)$

Determining Period.

$a)$The period $T$is given by

$T=\frac{1}{f}=\frac{1}{2}=0.5\mathrm{s}$

Finding angular frequency.

$b)$Angular frequency as a function in frequency is

$\omega =2\pi f\Rightarrow \left(1\right)$

Substitution in $\left(1\right)$yields

$\omega =2\pi \times 2=4\pi$

Finding Amplitude.

$c)$To get the amplitude we need to find the spring constant which is given by

$\omega =\sqrt{\frac{k}{m}}\Rightarrow k=m{\omega }^2=0.2\times (4\pi {)}^2=31.6\mathrm{N}/\mathrm{m}$

So use conservation's law

$\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2=\frac{1}{2}k{A}^2\Rightarrow \left(2\right)$

Use conditions $x\left(0\right)=0.05m\&v\left(0\right)=0.3m/s$to get that

$31.6(0.05{)}^2+0.2(0.3{)}^2=31.6A^2\Rightarrow A=0.055\mathrm{m}=5.5\mathrm{cm}$

To find phase constant.

$d)$To get phase constant we want to use initial condition $x\left(0\right)=5\mathrm{cm}$but let's start with equation

$x\left(t\right)=A\cos \left(\omega t+{\varphi }_o\right)\Rightarrow \left(3\right)$

Apply Condition to get that

$5=5.5\cos \left({\varphi }_o\right)\Rightarrow {\varphi }_o=\pm 0.137\pi$

The displacement is positive at $t=0so\cos$is positive and the angle may be in $1st$or $4th$quadratic where cos is positive, to determine which value s correct we need to determine it by comparing it with velocity .

Velocity is a sin wave which is positive in $1st$and $2nd$quadratic, negative in $3rd$and $4th$quadratic and we know that phase constant is in $1st$or $4th$quadratic so if the argument is in $1st$quadratic the argument of sin will be positive and the velocity will be negative because of - sign due to differentiation but if the argument is in Lth quadratic it will be negative and the velocity will be positive so the right answer is

${\varphi }_o=0.137\pi$

Finding maximum speed.

$e)$Speed is the differentiation of displacement with respect to time

$v\left(t\right)=-A\omega \sin \left(\omega t+{\varphi }_o\right)\Rightarrow \left(4\right)$

The max positive speed is

$v_{\max }=A\omega =4\pi \times 5.5=69\mathrm{cm}/\mathrm{s}$

To find maximum acceleration.

$f)$Acceleration is differentiation of speed with respect to times

$a\left(t\right)=-A{\omega }^2\cos \left(\omega t+{\varphi }_o\right)\Rightarrow \left(5\right)$

SO the maximum acceleration is given by

$a_{\max }=A{\omega }^2=5.5\times (4\pi {)}^2=869\mathrm{cm}·{\mathrm{s}}^{-2}$

To find Total energy.

$g)$The energy stored in the system is given by

$E=\frac{1}{2}k{A}^2=\frac{1}{2}\times 31.6\times (0.055{)}^2=0.05\mathrm{J}$

Finding the position When t=0.40 s.

$h)$Substitution in $\left(4\right)$yields

$v(0.4)=-69\sin (4\pi \times 0.4+0.137\pi )=50.74\mathrm{cm}/\mathrm{s}$