1. When we put
on spring a body with weight of 20 N and spring stretches for 40 cm. If we put
on same spring the body with weight of 10N the spring stretches for 35 cm.
Calculate the spring elasticity constant? What is the length of upstretched
spring?
$$\begin{align}
& {{G}_{1}}=20\text{ N} \\
& {{y}_{1}}+{{l}_{0}}=40\text{ cm}=0.4\text{ m} \\
& {{G}_{2}}=10\text{ N} \\
& \underline{{{y}_{2}}+{{l}_{0}}=35\text{ cm}=0.35\text{ m}} \\
& {{l}_{0}}=?,k=? \\
& {{y}_{1}}=0.4-{{l}_{0}} \\
& {{y}_{2}}=0.35-{{l}_{0}} \\
& k{{y}_{1}}={{G}_{1}} \\
& k{{y}_{2}}={{G}_{2}} \\
& \frac{{{G}_{1}}}{{{y}_{1}}}=\frac{{{G}_{2}}}{{{y}_{2}}} \\
& {{G}_{1}}{{y}_{2}}={{G}_{2}}{{y}_{1}} \\
& 20\left( 0.35-{{l}_{0}} \right)=10\left( 0.4-{{l}_{0}} \right) \\
& 7-20{{l}_{0}}=4-10{{l}_{0}} \\
& 10{{l}_{0}}=3 \\
& {{l}_{0}}=0.3\text{ m}=30\text{ cm} \\
& {{y}_{1}}=0.4-0.3=0.1\text{ m}=10\text{ cm} \\
& {{y}_{2}}=0.35-0.3=0.05\text{ m}=5\text{ cm} \\
& k{{y}_{1}}={{G}_{1}} \\
& k=\frac{{{G}_{1}}}{{{y}_{1}}}=\frac{20}{0.1}=200\text{ N/m} \\
\end{align}$$
2.The body hung on a spring vibrates harmonically with
circular frequency of 0.5 rad/s. Amplitude of harmonic vibrations is 0.8 m. Derive the equation which describes the relation between elongation y and time
t, speed u and time t and acceleration a and time t if at the moment t = 0 body
moves from equilibrium position in positive direction of y-axis.
$$\begin{align}
& \omega =0.5\text{ rad/s} \\
& \underline{A=0.8\text{ m }} \\
& y=?,v=?,a=? \\
& y=A\sin \left( \omega t+\theta \right) \\
& \theta =0 \\
& y=0.8sin\left( 0.5t \right) \\
& v=\frac{dy}{dt}=A\omega \cos \left( \omega t \right) \\
& v=0.8\cdot 0.5\cos \left( 0.5t \right) \\
& v=0.4\cos \left( 0.5t \right) \\
& a=\frac{dv}{dt}=\frac{{{d}^{2}}y}{d{{t}^{2}}}=-A{{\omega }^{2}}\sin \left( \omega t \right) \\
& a=-0.8\cdot {{0.5}^{2}}\sin \left( 0.5t \right) \\
& a=-0.2\sin \left( 0.5t \right) \\
\end{align}$$
3. If the body
vibrates harmonically with amplitude of 4 cm determine the distance that body
travels during one period.
$$\begin{align}
& A=4\text{ cm}=0.04\text{ m} \\
& \underline{\text{t=T }} \\
& s=4A=4\cdot 0.04=0.16\text{ m} \\
& s=16\text{ cm} \\
\end{align}$$
4. Equation
of the body with harmonic oscillations is given by:
$$y=0.02\sin \left( \frac{\pi }{2}t+\frac{\pi }{4} \right)$$
a)Amplitude.From
given equation the amplitude is:
$$A=0.02\text{ m}=2\text{ cm}$$
b)Circular
frequency and period. From
given equation the circular frequency is:
$$\begin{align}
& \omega =\frac{\pi }{2} \\
& T=\frac{2\pi }{\omega }=\frac{2\pi }{\frac{\pi }{2}}=4s \\
\end{align}$$
c) Maximum speed. To find the maximum speed first
we need to derive the equation of speed and then plot it in order to visualize
$$\begin{align}
& y=0.02\sin \left( \frac{\pi }{2}t+\frac{\pi }{4} \right) \\
& u=\frac{dy}{dt}=0.02\frac{\pi }{2}\cos \left( \frac{\pi }{2}t+\frac{\pi }{4} \right) \\
& u=0.0314159\cos \left( \frac{\pi }{2}t+\frac{\pi }{4} \right) \\
\end{align}$$
The
u,t graph is shown in the next figure.
$$\begin{align}
& u=0.0314159\cos \left( \frac{\pi }{2}t+\frac{\pi }{4} \right) \\
& 0.0314159\cos \left( \frac{\pi }{2}t+\frac{\pi }{4} \right)=0.0314159/:0.0314159 \\
& \cos \left( \frac{\pi }{2}t+\frac{\pi }{4} \right)={1}/{\arccos }\; \\
& \frac{\pi }{2}t+\frac{\pi }{4}={2\pi n}/{4}\; \\
& 2\pi t+\pi =8\pi n \\
& 2\pi t=8\pi n-\pi \\
& \text{for n=1} \\
& 2\pi t=7\pi \\
& t=\frac{7}{2}=3.5s \\
& \text{Now let }\!\!'\!\!\text{ s get back into the equation of speed and substitute time t}\text{. } \\
& u=0.0314159\cos \left( \frac{\pi }{2}\cdot \frac{7}{2}+\frac{\pi }{4} \right) \\
& u=0.0314159\cos \left( \frac{7\pi }{4}+\frac{\pi }{4} \right) \\
& u=0.0314159\cos \left( \frac{8\pi }{4} \right)=0.0314159\cos \left( 2\pi \right)=0.0314159\text{ m/s} \\
& u=3,14159\text{ cm/s} \\
\end{align}$$
d)Maximum acceleration. In order to obtain the
maximum acceleration first we need to obtain the expression for acceleration in
other words we need to derivate expression for speed
$$\begin{align}
& u=0.0314159\cos \left( \frac{\pi }{2}t+\frac{\pi }{4} \right) \\
& a=\frac{du}{dt}=-0.0314159\frac{\pi }{2}\sin \left( \frac{\pi }{2}t+\frac{\pi }{4} \right) \\
& a=-0.04934798\sin \left( \frac{\pi }{2}t+\frac{\pi }{4} \right) \\
& t=? \\
& -0.04934798\sin \left( \frac{\pi }{2}t+\frac{\pi }{4} \right)={-0.04934798}/{\left( -0.04934798 \right)}\; \\
& \sin \left( \frac{\pi }{2}t+\frac{\pi }{4} \right)={1}/{\arcsin }\; \\
& \frac{\pi }{2}t+\frac{\pi }{4}=\frac{\pi }{2}+2\pi n \\
& \text{If n=0} \\
& \frac{\pi }{2}t+\frac{\pi }{4}={\frac{\pi }{2}}/{4}\; \\
& 2\pi t+\pi =2\pi \\
& 2\pi t={\pi }/{2\pi }\; \\
& t=\frac{1}{2}=0.5s \\
& a=-0.04934798\sin \left( \frac{\pi }{2}t+\frac{\pi }{4} \right)=0.04934798m/s \\
& a=4.9347cm/s \\
\end{align}$$
e)Initial
phase angle
$${{\theta }_{0}}=\frac{\pi }{4}$$
5. Elongation
of the body which harmonically oscillates in time is shown in next figure. Determine
the oscillation amplitude, period, circular frequency and phase angle. Write
down the equation of harmonic oscillation motion, speed and acceleration.
$$\begin{align}
& A=4m \\
& \underline{T=4s} \\
& \omega =\frac{2\pi }{T}=\frac{2\pi }{4}=\frac{\pi }{2}rad/s \\
& {{\theta }_{0}}=\pi \\
& y=A\sin \left( \omega t+{{\theta }_{0}} \right) \\
& y=4\sin \left( \frac{\pi }{2}t+\pi \right) \\
& u=A\omega \cos \left( \omega t+{{\theta }_{0}} \right) \\
& u=4\frac{\pi }{2}\cos \left( \frac{\pi }{2}t+\pi \right) \\
& u=2\pi \cos \left( \frac{\pi }{2}t+\pi \right) \\
& a=-A{{\omega }^{2}}\sin \left( \frac{\pi }{2}t+\pi \right) \\
& a=-4\frac{{{\pi }^{2}}}{4}\sin \left( \frac{\pi }{2}t+\pi \right) \\
& a=-{{\pi }^{2}}\sin \left( \frac{\pi }{2}t+\pi \right) \\
\end{align}$$
6. Figure
shows body with mass m attached on the spring. The spring is stretched for 5 cm
and then we let her go and the body oscillates with period of 2 s.
Determine:
a)The
equation of oscillating body if at the time t= 0 body was in position B and was
moving towards the position A.
b)The
equation of oscillating body if at the moment t= 0 body was at position A.
c)The
equation of oscillating body if at the moment t= 0 body was at position C.
d)Highest
speed is at position?
e)Highest
acceleration is at position?
f)Kinetic
energy of the body is highest at position?
g)Potential
energy is highest at position?
Solution
$$\begin{align}
& A=5cm=0.05m \\
& T=2s \\
& \underline{\omega =\frac{2\pi }{T}=\pi rad/s} \\
& a)y=f(t)=?\text{ }t=0s\text{ at point B moving to point A} \\
& b)y=f(t)=?\text{ t=0s at point A} \\
& c)y=f(t)=?\text{ t=0s at point C} \\
& \text{d)Maximum speed} \\
& \text{e)Maximum acceleration} \\
& \text{f)Kinetic energy maximum?} \\
& \text{g)Potential energy?} \\
\end{align}$$
a) The
equation of oscillating body if at the time t= 0 body was in position B and was
moving towards the position A.
$$\begin{align}
& y=A\sin \left( \omega t+{{\theta }_{0}} \right) \\
& y=0.05\sin \left( \pi t \right) \\
& t=0s \\
& y=0m \\
\end{align}$$
b)The
equation of oscillating body if at the moment t= 0 body was at position A.
$$\begin{align}
& y=A\sin \left( \omega t+{{\theta }_{0}} \right) \\
& y=0.05\sin \left( \pi t+\frac{\pi }{2} \right) \\
& t=0s \\
& y=0.05m \\
\end{align}$$
c)The
equation of oscillating body if at the moment t = 0 body was at position C.
$$\begin{align}
& y=A\sin \left( \omega t+{{\theta }_{0}} \right) \\
& y=0.05\sin \left( \pi t+\frac{3\pi }{2} \right) \\
& t=0s \\
& y=-0.05m \\
\end{align}$$
d)Highest
speed is at position?
$$\begin{align}
& y=0.05\sin \left( \pi t \right) \\
& u=\frac{dy}{dt}=0.05\pi \cos \left( \pi t \right) \\
& u=0,157\cos \left( \pi t \right) \\
& 0,157\cos \left( \pi t \right)=0,157 \\
& \cos \left( \pi t \right)=1 \\
& \pi t=2\pi n \\
& n=0 \\
& t=0s \\
& u=0,157\cos \left( 0 \right)=0,157m/s \\
\end{align}$$
Maximum
speed is at positon A, and C.
e)Maximum
acceleration
$$\begin{align}
& u=\frac{dy}{dt}=0.05\pi \cos \left( \pi t \right) \\
& a=-0.05{{\pi }^{2}}\sin \left( \pi t \right) \\
& a=-0.49\sin \left( \pi t \right) \\
\end{align}$$
$$\begin{align}
& a=-0.49\sin \left( \pi t \right) \\
& -0.49\sin \left( \pi t \right)=0.49 \\
& \sin \left( \pi t \right)=1 \\
& \pi t=\frac{3\pi }{2} \\
& t=\frac{3}{2}=1.5s \\
& a=-0.49\sin \left( \frac{3}{2}\pi \right)=0.49m/{{s}^{2}} \\
\end{align}$$
f)Maximum
kinetic energy of the system
$$\begin{align}
& {{E}_{k}}=\frac{m{{\omega }^{2}}}{2}y_{0}^{2}{{\cos }^{2}}\left( \omega t+\theta \right) \\
& \text{Assumption }m=1kg \\
& {{E}_{k}}=0.012337J \\
\end{align}$$
The maximum kinetic energy is at
point B.
g)Potential energy is highest at position?
$$\begin{align}
& {{E}_{p}}=\frac{m{{\omega }^{2}}}{2}y_{0}^{2}{{\sin }^{2}}\left( \omega t+\theta \right) \\
& \text{Assumption }m=1kg \\
& {{E}_{p}}=0.00125J \\
\end{align}$$
Maximum potential energy is highest at position A and C.
7.Body of
mass 0.4 kg is attached on spring and oscillates around equilibrium position
with frequency 2 Hz and amplitude 20 cm.
a) Derive
the equation that describes the oscillation of the body if at the moment t= 0
the body was at equilibrium position and was moving in positive direction of
y-axis.
b) Derive
the equation that describes the motion of the body if at the moment t = 0 the
body was 20 cm from equilibrium position in negative direction of y-axis.
c) Determine
the speed of the moving body when it goes through the equilibrium position
d) Derive
the equation for body oscillation speed in dependence of time for cases a and b
and draw v,t graphs
e) Determine
the speed of the body when the body is 10 cm from equilibrium position.
f) Derive
the equation for acceleration of the moving body for cases a) and b) and draw a,t graphs
g) Determine
the total energy of the system
h) Determine
the maximum force that is acting on the body
i)
In which position body has maximum speed and in which
maximum acceleration. What is the ratio between acceleration and the speed in
each moment? Does this ratio changes over time ?
j)
What is the ratio between acceleration and elongation
in each moment?
k) Draw
a graph which shows how speed changes over elongation u =f(y).
Solution:
$$\begin{align}
& m=0.4kg \\
& A=20cm=0.2m \\
& f=2Hz \\
& T=1/f=1/2s \\
& \underline{\omega =\frac{2\pi }{T}=4\pi rad/s} \\
& a)y=f(t)=?\text{ }t=0s\text{ yt-graph} \\
& b)y=f(t)=?\text{ t=0s at point A!!!! ovo sve treba izmjeniti} \\
& c)y=f(t)=?\text{ t=0s at point C} \\
& \text{d)Maximum speed} \\
& \text{e)Maximum acceleration} \\
& \text{f)Kinetic energy maximum?} \\
& \text{g)Potential energy?} \\
\end{align}$$
a)Derive the equation that describes the oscillation of the body if at the moment t= 0 the body was at equilibrium position and was moving in positive direction of y-axis.
$$\begin{align}
& y=A\sin \left( \omega t+{{\theta }_{0}} \right) \\
& y=0.2\sin \left( 4\pi t \right) \\
\end{align}$$
b)Derive the equation that describes the motion of the body if at the moment t = 0 the body was 20cm from equilibrium position in negative direction of y-axis.
$$\begin{align}
& y=A\sin \left( \omega t+{{\theta }_{0}} \right) \\
& y=0.2\sin \left( 4\pi t-\frac{\pi }{2} \right) \\
\end{align}$$
$$\begin{align}
& y=0.2\sin \left( 4\pi t \right) \\
& u=\frac{dy}{dt}=0.2\cdot 4\pi \cos \left( 4\pi t \right) \\
& u=2.51\cos \left( 4\pi t \right) \\
& t=0s \\
& u=2.51m/s \\
\end{align}$$
d)Derive the equation for body oscillation speed in dependence of time for cases a and b and draw u,t graphs
For the case a the speed of body oscillation is:
$$u=2.51\cos \left( 4\pi t \right)$$
For the case b the equation which describes the body oscillation speed can be written in the following form:
$$\begin{align}
& y=0.2\sin \left( 4\pi t-\frac{\pi }{2} \right) \\
& u=\frac{dy}{dt}=2.51\cos \left( 4\pi t-\frac{\pi }{2} \right) \\
\end{align}$$
e)Determine the speed of the body when the body is 10 cm from equilibrium position.
$$\begin{align}
& y=0.2\sin \left( 4\pi t \right) \\
& 0.2\sin \left( 4\pi t \right)=0.1 \\
& \sin \left( 4\pi t \right)=0.5 \\
& 4\pi t=\frac{\pi }{6} \\
& t=\frac{1}{24}=0.0416667s \\
& u=2.51cos\left( 4\pi t \right) \\
& u=2.51\cos \left( \frac{\pi }{6} \right)=2.173m/s \\
\end{align}$$
f)Derive the equation for acceleration of the moving body for cases a) and b) and draw a,t graphs
For the case a the equation of acceleration:
$$\begin{align}
& u=2.51cos\left( 4\pi t \right) \\
& a=\frac{du}{dt}=-2.51\cdot 4\cdot \pi sin\left( 4\pi t \right) \\
& a=-31.5159\sin \left( 4\pi t \right) \\
\end{align}$$
For the case b the equation of acceleration:
$$\begin{align}
& u=2.51\cos \left( 4\pi t-\frac{\pi }{2} \right) \\
& a=\frac{du}{dt}=-2.51\cdot 4\cdot \pi sin\left( 4\pi t-\frac{\pi }{2} \right) \\
& a=-31.5159\sin \left( 4\pi t-\frac{\pi }{2} \right) \\
\end{align}$$
8.The amplitude of harmonic oscillation is 30cm and
period 2s. In moment t= 0s elongation is 0 and the body moves in positive
direction of y-axis. Determine the elongation, speed and acceleration at the
moment t=T/12.
$$\begin{align}
& A=30cm=0.3m \\
& T=2s \\
& \underline{t=T/12=1/6s} \\
& y=? \\
& v=? \\
& a=? \\
& y=A\sin \left( \omega t \right) \\
& \omega =\frac{2\pi }{T}=\frac{2\pi }{2}=\pi \text{ rad/s}\ \\
& y=0.3\sin \left( \pi t \right)=0.3\sin \left( \frac{\pi }{6} \right)=0.15\text{ m} \\
& y=15\text{ cm} \\
& v=A\omega cos\left( \omega t \right)=0.942478\cos \left( \pi t \right) \\
& v=0.942478\cos \left( \frac{\pi }{6} \right)=0.8162\text{ m/s} \\
& v=81.62\text{ cm/s} \\
& a=-2.96088\sin \left( \omega t \right)=-2.96088\sin \left( \frac{\pi }{6} \right) \\
& a=-1.48044\text{ m/s }=-148.044\text{ cm/s} \\
& a=\left| -148.044 \right|=148.044\text{ cm/s} \\
\end{align}$$
9.The body is
attached to a spring and is moved from equilibrium position for 10cm than is
left to oscillate. At what distance from the equilibrium position will:
a) Speed
of the body be equal to one half of maximum speed?
b)Acceleration
of the body be equal to one half of maximum acceleration?
$$\begin{align}
& \underline{A=10cm=0.1m} \\
& a)v=0.5{{v}_{\max }} \\
& y=? \\
& y=A\sin \left( \omega t \right) \\
& v=A\omega \cos \left( \omega t \right) \\
& v={{v}_{\max }}\Rightarrow \cos \left( \omega t \right)=1 \\
& \omega t=\alpha =2n\pi \\
& n=0\Rightarrow \alpha =0 \\
& v=0.5{{v}_{\max }}\Rightarrow \cos \left( \omega t \right)=0.5 \\
& \omega t=\alpha =\frac{\pi }{3} \\
& y=A\sin \left( \omega t \right)=0.1\cdot \sin \left( 60{}^\circ \right)=0.0866\text{ m} \\
& y=8.66\text{ cm} \\
& b)a=0.5{{a}_{\max }} \\
& a={{a}_{\max }}\Rightarrow \sin \left( \omega t \right)=1 \\
& \omega t=\alpha =\frac{n\pi }{2} \\
& n=1\Rightarrow \alpha =\frac{\pi }{2} \\
& a=0.5{{a}_{\max }}\Rightarrow \sin \left( \omega t \right)=0.5 \\
& \omega t=\alpha =\frac{\pi }{6} \\
& y=A\sin \left( \omega t \right)=0.1\cdot 0.5=0.05\text{ m} \\
& y=5\text{ cm} \\
\end{align}$$
10.The body
harmonically oscillates with the circular frequency of 0.5 rad/s. Amplitude of
the oscillations is 80 cm.
a)Derive
the equations of elongation, speed and acceleration if at the moment t=0 the
body was in equilibrium position and was moving in positive direction of
y-axis.
b)Derive
the equation that connects speed and elongation.
$$\begin{align}
& \omega =0.5rad/s \\
& \underline{A=80cm\text{ }} \\
& a)y=A\sin \left( \omega t \right) \\
& y=0.8\sin \left( 0.5t \right) \\
& u=\frac{dy}{dt}=A\omega \cos \left( \omega t \right) \\
& u=0.4\cos \left( 0.5t \right) \\
& a=\frac{du}{dt}=-A{{\omega }^{2}}\sin \left( \omega t \right) \\
& a=-0.2sin\left( 0.5t \right) \\
& b)y=A\sin \left( \omega t \right)\Rightarrow \sin \left( \omega t \right)=\frac{y}{A} \\
& v=A\omega \cos \left( \omega t \right)\Rightarrow \cos \left( \omega t \right)=\frac{v}{A\omega } \\
& {{\sin }^{2}}\left( \omega t \right)+{{\cos }^{2}}\left( \omega t \right)=1 \\
& \frac{{{y}^{2}}}{{{A}^{2}}}+\frac{{{v}^{2}}}{{{A}^{2}}{{\omega }^{2}}}={1}/{{{A}^{2}}{{\omega }^{2}}}\; \\
& {{y}^{2}}{{\omega }^{2}}+{{v}^{2}}={{A}^{2}}{{\omega }^{2}} \\
& {{v}^{2}}={{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right) \\
& {{v}_{1,2}}=\pm \text{ }\omega \text{ }\sqrt{{{A}^{2}}-{{y}^{2}}} \\
& {{v}_{1,2}}=\pm \text{ }0.5\text{ }\sqrt{0.64-{{y}^{2}}} \\
\end{align}$$
11.The period
of harmonic oscillations is 3.6 s determine the shortest time what is needed
for the body which oscillates to distance itself from equilibrium position for
half amplitude.
$$\begin{align}
& T=3.6\text{ s} \\
& \underline{y=0.5A} \\
& t=? \\
& \omega =2\pi f=\frac{2\pi }{T}=\frac{2\pi }{3.6}=1.74\text{ rad/s} \\
& y=A\sin \left( \omega t \right) \\
& 0.5A=A\sin \left( 1.74t \right) \\
& 0.5=\sin \left( 1.74t \right) \\
& 1.74t=\arcsin \left( 0.5 \right) \\
& 1.74t=\frac{\pi }{6} \\
& t=\frac{\pi }{6\cdot 1.74}=0.3\text{ s} \\
\end{align}$$
12.Body of mass 1 kg harmonically oscillates according to
the equation y = 0.32 sin(7.4 t). Determine the:
a) amplitude,
b) frequency and
c) kinetic
and potential energy of the body when the body is 0.26 m from equilibrium
position.
$$\begin{align}
& m=1\text{ kg} \\
& \underline{y=0.32\sin \left( 7.4t \right)} \\
& a)A=? \\
& A=0.32\text{ }m \\
& b)f=? \\
& \omega =2\pi f \\
& f=\frac{\omega }{2\pi }=\frac{7.4}{2\pi }=1.18\text{ Hz} \\
& c)y=0.26\text{ m} \\
& {{E}_{k}}=? \\
& {{E}_{p}}=? \\
& y=0.32\sin \left( 7.4t \right) \\
& \frac{0.26}{0.32}=\sin \left( 7.4t \right) \\
& 7.4t=\arcsin \left( 0.8125 \right) \\
& t=\frac{54.34}{7.4}=7.34s \\
& {{E}_{k}}=\frac{m{{\omega }^{2}}}{2}{{A}^{2}}{{\cos }^{2}}\left( \omega t+\theta \right) \\
& {{E}_{k}}=\frac{1\cdot {{7.4}^{2}}}{2}{{0.32}^{2}}{{\cos }^{2}}\left( 7.4\cdot 7.34 \right)=1.059\text{ J} \\
& {{E}_{p}}=\frac{m{{\omega }^{2}}}{2}{{A}^{2}}{{\sin }^{2}}\left( \omega t+\theta \right)=\frac{1\cdot {{7.4}^{2}}}{2}{{0.32}^{2}}{{\sin }^{2}}\left( 7.4\cdot 7.34 \right) \\
& {{E}_{p}}==1.74468\text{ J} \\
\end{align}$$
13.What
variables in harmonic oscillations in amplitude position have the maximum value?
Elastic energy of a spring in dependence on time is shown in next figure.
Elastic energy in dependence on time is shown in following figure.
From previous figures we see that the system at amplitude position has the maximum energy.
The following figures show elongation, speed and acceleration in dependence of time.
As we can see from previous figure the maximum value at amplitude positions is acceleration.
The dependence of potential and kinetic energy of time is show in following figure.
From previous figures the maximum values at amplitude position are Elastic force, Potential Energy and Acceleration.
14. Body
of mass 1 kg is attached to the horizontally laid spring with spring constant
of 120 N/m. At the moment t = 0 the body is moved so spring is compressed. The
initial speed of the body is 3 m/s. If friction between body and a surface is
neglected determine:
a)Period
and frequency of oscillations
b)Amplitude
c)Maximum
acceleration
d)Total
energy
$$\begin{align}
& m=1\text{ kg} \\
& \underline{\begin{align}
& k=120N/m \\
& t=0\text{ s} \\
& v=3m/s \\
\end{align}} \\
& a)T=? \\
& f=? \\
& T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{1}{120}}=0.57\text{ s} \\
& f=\frac{1}{T}=\frac{1}{0.57}=1.74346\text{ s} \\
& b)A=? \\
& \omega =2\pi f=10.9545\text{ rad/s} \\
& v=A\omega \cos \left( \omega t \right) \\
& A=\frac{v}{\omega \cos \left( \omega t \right)}=\frac{3}{10.9545\cdot \cos \left( 10.9545\cdot 0 \right)} \\
& A=\frac{3}{10.9545}=0.27386\text{ m} \\
& c){{a}_{\max }}=? \\
& a={{a}_{\max }}\Rightarrow y=A\Rightarrow t=\frac{T}{4},\frac{3T}{4} \\
& t=0.143393\text{ s} \\
& a=-A{{\omega }^{2}}\sin \left( \omega t \right) \\
& a=-0.27386\cdot {{10.9545}^{2}}\sin \left( \frac{10.9545\cdot 0.143393\cdot 180}{\pi } \right) \\
& a={{a}_{\max }}=32.86\text{ m/}{{\text{s}}^{2}} \\
& d)E=\frac{m{{\omega }^{2}}}{2}{{A}^{2}}=\frac{1\cdot {{10.9545}^{2}}}{2}\cdot {{0.27386}^{2}}=4.5\text{ J} \\
\end{align}$$
15.The
body of mass 0.4 kg is attached on spring and oscillates with frequency of 3
Hz. Determine the frequency of oscillations if the mass of the body is 0.1 kg?
$$\begin{align}
& {{m}_{1}}=0.4\text{ kg} \\
& \underline{\begin{align}
& {{f}_{1}}=3\text{ Hz} \\
& {{m}_{2}}=0.1\text{ kg} \\
\end{align}} \\
& {{f}_{2}}=? \\
& {{k}_{1}}={{k}_{2}}=k \\
& {{T}_{1}}=\frac{1}{{{f}_{1}}} \\
& {{T}_{1}}=2\pi \sqrt{\frac{{{m}_{1}}}{k}} \\
& \frac{1}{{{f}_{1}}}=2\pi \sqrt{\frac{{{m}_{1}}}{k}}\Rightarrow k=4f_{1}^{2}{{m}_{1}}{{\pi }^{2}} \\
& k=4\cdot {{3}^{2}}\cdot 0.4\cdot {{\pi }^{2}}=142.122\text{ N/m} \\
& {{f}_{2}}=\frac{1}{2\pi \sqrt{\frac{{{m}_{2}}}{k}}}=\frac{1}{2\pi \sqrt{\frac{0.1}{142.122}}}=6\text{ Hz} \\
\end{align}$$
16.Elastic
spring is transferred to a planet where gravitational acceleration is equal to
one half of gravitational acceleration on earth. The period of spring
oscillation when compared to period oscillation on Earth will be:
a)Unchanged
b)Two
times bigger
c)Two
times smaller
d)Four
times bigger
e)Four
times smaller
$$\begin{align}
& \underline{{{g}_{p}}=0.5\cdot {{g}_{earth}}} \\
& \frac{{{T}_{earth}}}{{{T}_{P}}}=? \\
& T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{G}{gk}} \\
& \frac{{{T}_{earth}}}{{{T}_{P}}}=\frac{\sqrt{\frac{{{G}_{earth}}}{{{g}_{earth}}}}}{\sqrt{\frac{{{G}_{p}}}{{{g}_{p}}}}}=\sqrt{\frac{\frac{{{G}_{earth}}}{{{g}_{earth}}}}{\frac{{{G}_{p}}}{0.5{{g}_{earth}}}}}=\sqrt{\frac{0.5{{G}_{earth}}}{{{G}_{p}}}} \\
& {{m}_{earth}}={{m}_{p}} \\
& \frac{{{G}_{earth}}}{{{g}_{earth}}}=\frac{{{G}_{p}}}{{{g}_{p}}}\Rightarrow \frac{{{G}_{earth}}}{{{g}_{earth}}}=\frac{{{G}_{p}}}{0.5\cdot {{g}_{earth}}} \\
& {{G}_{p}}=0.5{{G}_{earth}} \\
& \frac{{{T}_{earth}}}{{{T}_{P}}}=\sqrt{\frac{0.5{{G}_{earth}}}{{{G}_{p}}}}=\sqrt{\frac{0.5{{G}_{earth}}}{0.5{{G}_{earth}}}}=1 \\
& {{T}_{earth}}={{T}_{p}} \\
\end{align}$$
Answer: the periods of oscillations are the same or unchanged.
17.Spring has attached weight and the system
oscillates and makes 45 oscillations per minute. What to do with the mass of a
weight so the system would oscillate with 30 oscillations per minute ?
