CBSE Questions for Class 11 Engineering Physics Oscillations Quiz 3 - MCQExams.com

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 force and acceleration is always in opposite direction of displacement. so the dot product of displacement $$\vec r$$ with $$\vec F$$ and $$\vec a$$ is always smaller than zero. $$\Rightarrow$$ B and D are correct 
• Force and acceleration are always in same direction so dot product of $$\vec F$$ and $$\vec a$$ is always greater then zero. $$\Rightarrow$$ A is incorrect
• Velocity and displacement are not always in the same direction so dot product of
  
  $$\vec v$$ and $$\vec r$$ is sometimes greater then zero and sometimes less then zero $$\Rightarrow$$ C is incorrect
  

$$x=A\sin { \omega t } \\ \Rightarrow \dfrac { dx }{ dt } =A\omega \cos { \omega t } \\ \Rightarrow v=A\omega \cos { \omega t } $$

$$v=\dfrac { v_{ max } }{ 2 } =\dfrac { A\omega  }{ 2 } $$

$$\Rightarrow \dfrac { A\omega  }{ 2 } =A\omega \cos { \omega t } \\ \Rightarrow \cos { \omega t } =\dfrac { 1 }{ 2 } \\ \Rightarrow \sin { \omega t } =\sqrt { 1-{ \left( \dfrac { 1 }{ 2 }  \right)  }^{ 2 } } =\sqrt { 1-{ \dfrac { 1 }{ 4 }  } } =\dfrac { \sqrt { 3 }  }{ 2 } \\ \Rightarrow x=a\sin { \omega t } =\dfrac { a\sqrt { 3 }  }{ 2 } $$

HINT: Circular motion is not oscillatory motion.

Step 1: Explanation
In circular motion because in the circular motion the particle does not move to and fro about a mean position as it is required for the SHM.

$$\textbf{Correct option: Option (A).}$$

$$\textbf{Explanation:}$$

The motion of particle moving in a circular path with uniform speed is periodic and is not oscillatory because it’s not to and fro. Hence it can also not be SHM and angular SHM.

$$\textbf{Thus, option (A) is only correct.}$$

$$\textbf{Correct option: Option (A).}$$

$$\textbf{Explanation:}$$

(A) A simple harmonic motion is always repeated at a regular interval and hence it is necessarily periodic.

(B) An SHM is a to and fro motion and hence it is necessarily oscillatory and hence this statement is incorrect.

(C) An oscillatory motion need not be always periodic and hence this statement is incorrect.

(D) A periodic motion may or may not be oscillatory. So this statement is also false.

$$\textbf{Thus, only option (A) is correct.}$$