Question 310: Oscillations

Question

A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is $$50 \%$$ of total energy?
$$\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)$$

Accepted Answer

The total energy is given by $$T \cdot E=\frac{1}{2} \mathrm{kA}^2$$
Kinetic energy is given by $$\mathrm{K} \cdot \mathrm{E}=\frac{1}{2} \mathrm{k}\left(\mathrm{A}^2-\mathrm{x}^2ight)$$
$$\begin{array}{ll} 
& 	ext { K.E }=\frac{1}{2} P . E \quad 	ext{...(given)}\
	herefore \quad & \frac{1}{2} k\left(A^2-x^2ight)=\frac{1}{2}\left(\frac{1}{2} k A^2ight) \
	herefore \quad & \left(A^2-x^2ight)=\frac{1}{2} A^2 \
	herefore \quad & \left(A^2-x^2ight)=\frac{1}{2} A^2 \
	herefore \quad & x=\frac{A}{\sqrt{2}}
\end{array}$$
The equation of displacement in SHM is
$$\begin{aligned}
& \mathrm{x}=\mathrm{A} \sin \frac{2 \pi \mathrm{t}}{\mathrm{T}} \
& 	herefore \quad \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A} \sin \frac{2 \pi \mathrm{t}}{4} \quad ....(\because 	ext{T}=	ext{4 sec})\
& 	herefore \quad \frac{1}{\sqrt{2}}=\sin \frac{\pi \mathrm{t}}{2} \
& 	herefore \quad \frac{\pi \mathrm{t}}{2}=\sin ^{-1} \frac{1}{\sqrt{2}} \
& \quad \frac{\pi \mathrm{t}}{2}=\frac{\pi}{4} \
& 	herefore \quad \mathrm{t}=0.5 \mathrm{~s}
\end{aligned}$$

Answer

0.1 s

Answer

0.2 s

Answer

0.4 s

Answer

0.5 s

Oscillations

A particle starts from mean position and performs S.H.M. with period 4 second. At what time its kinetic energy is $50 \% of total energy? \left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)

Choose the correct answer: