Oscillations Examples I

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.

$$\begin{align} & y=A\sin \left( \omega t+{{\theta }_{0}} \right) \\ & y=0.2\sin \left( 4\pi t-\frac{\pi }{2} \right) \\ \end{align}$$

c) Determine the speed of the moving body when it goes through the equilibrium position

  $$\begin{align} & y=0.2\sin \left( 4\pi t \right) \\ & u=\frac{dy}{dt}=0.2\cdot 4\pi \cos \left( 4\pi t \right) \\ & u=2.51\cos \left( 4\pi t \right) \\ & t=0s \\ & u=2.51m/s \\ \end{align}$$

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 ?