Oscillations Test - 18

$$x = A sin\  \omega t$$
$$V = \dfrac {dx}{dt} = A\omega\ cos \omega t$$
$$\dfrac {^\dfrac {3T}{8}_{0} \int V dt}{^\dfrac {3T}{8}_{0} \int dt} = \dfrac {\dfrac {A\omega}{\omega}\times [sin \omega t]^{\dfrac {3T}{8}}_{0}}{[t]^{\dfrac {3T}{8}}_{0}}$$

$$x=asin^{2}t$$

Since a SHM can be represented by $$x=A\sin { \left( \omega t+\phi  \right)  } $$

$$\Rightarrow x=A\sin { \left( \left( \dfrac { 2\pi  }{ T }  \right) t+\phi  \right)  } $$ ; $$T=6s$$

so the equation becomes
$$x=A\sin { \left( \dfrac { \pi t }{ 3 } +\phi  \right)  } $$

We are given that at $$t=1s, x=0 $$; so we have
$$0=A\sin { \left( \dfrac { \pi  }{ 3 } +\phi  \right)  } \\ \Rightarrow \left( \dfrac { \pi  }{ 3 } +\phi  \right) =0\\ \Rightarrow \phi =-\dfrac { \pi  }{ 3 } \\ \Rightarrow x=A\sin { \left( \dfrac { \pi t }{ 3 } -\dfrac { \pi  }{ 3 }  \right)  } \\ \Rightarrow v=\dfrac { dx }{ dt } =\dfrac { d }{ dt } \left( A\sin { \left( \dfrac { \pi t }{ 3 } -\dfrac { \pi  }{ 3 }  \right)  }  \right) =\dfrac { \pi A }{ 3 } \cos { \left( \dfrac { \pi t }{ 3 } -\dfrac { \pi  }{ 3 }  \right)  }   $$
Now we are given that at $$t=2s, \left| v \right| =0.25m/s$$, so we have
$$0.25=\left| \dfrac { \pi A }{ 3 } \cos { \left( \dfrac { 2\pi  }{ 3 } -\dfrac { \pi  }{ 3 }  \right)  }  \right| \\ \Rightarrow 0.25=\left| \dfrac { \pi A }{ 3 } \cos { \left( \dfrac { \pi  }{ 3 }  \right)  }  \right| \\ \Rightarrow 0.25=\dfrac { \pi A }{ 6 } \\ \Rightarrow A=\dfrac { 3 }{ 2\pi  } $$

For small oscillations about the positions of stable equilibrium, let $$\mathrm{x}=-\mathrm{a}+\mathrm{x}'$$ where $${\times }'< < a.$$ Then the equation of motion becomes

$$m\frac{d^{2}{\times }'}{dt^{2}}=\frac{-C{\times }'(2a-{\times }')}{\left [ ({\times }'-a)^{2}+a^{2} \right ]^{2}}$$

$$\simeq -\frac{C{\times }'}{2a^{3}}$$

About x = - a : period T = $$\frac{2\pi }{\omega }=2\pi \sqrt{\frac{2ma^{3}}{C}}$$

$$v_{max}=a\omega$$
$$v=n\lambda $$
$$\dfrac{v_{max}}{v}=\dfrac{a\omega}{n\lambda }=\dfrac{a(2\pi n)}{n\lambda}=\dfrac{2\pi a}{\lambda } =\dfrac{2\pi a}{\dfrac{2\pi }{K}}$$
$$=Ka=\dfrac{\pi }{2}\times 3=\dfrac{3\pi }{2}$$

As seen from graph, we can infer that for $$x<0$$ particle is in SHM as in spring. and for $$x>0$$ particle is in influence of gravity i.e. thrown upwards with some initial velocity.  
So we can divide the time period in two parts. first $$T_1$$ and second $$T_2$$
$$T_1$$ is half of the time period of a full SHM i.e. $$T_1=\pi \sqrt{\dfrac{m}{k}}$$
$$E=\dfrac{1}{2}m{v_{max}}^2 \\ \Rightarrow v_{max}=\sqrt{\dfrac{2E}{m}}$$
and $$T_2$$ is the time during which the particle remains in air when it is thrown upwards with velocity $$v_{max}=\sqrt{\dfrac{2E}{m}}$$
$$0=\sqrt{\dfrac{2E}{m}} t - \dfrac{1}{2}gt^2$$            Since $$s=ut - \dfrac{1}{2}gt^2$$
$$\Rightarrow T_2={2\sqrt{2E/mg^2}}$$
So total time time period $$T=\pi \sqrt{\dfrac{m}{k}}+{2\sqrt{\dfrac{2E}{mg^2}}}$$

Since it's speed is continuously increasing, it will not cover same distance in same time interval. there is no restoring force acting on it. there is no mean position. So it can not be periodic, oscillatory and SHM.

In simple harmonic motion, displacement of a particle is directly proportional to restoring force.
$$F \propto -x \Rightarrow F=-kx \Rightarrow m\dfrac{d^2x}{dt^2}=-kx$$.
The solution of the above equation can be given by
$$x=c_1\cos(\omega t)+c_2\sin(\omega t) = A \sin (\omega t -\phi)$$, where
$$A=\sqrt{c_1^2+c_2^2}$$, $$\omega=\sqrt{\dfrac{k}{m}}$$ and $$\phi =\tan^{-1}(\dfrac{c_2}{c_1})$$
$$c_1$$ and $$c_2$$ are constants and values can be determined from the initial boundary conditions.
So simple harmonic motions can be described as sin or cosine functions.

Ans: $$A$$

Answer: 1
For Simple Harmonic Motion, 
$$a(t)=-\omega^2 x(t)$$
Differentiate with respect to x.
$$\frac{da}{dx}=-\omega^2\frac{dx}{dx}=-\omega^2$$
From graph given the slope is $$-\frac{\beta}{\alpha}$$
Therefore, $$\frac{da}{dx}=-\frac{\beta}{\alpha}$$
$$\omega=\sqrt{\frac{\beta}{\alpha}}$$
Finally, frequency of oscillation, $$f=\frac{\omega}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{\beta}{\alpha}}$$