Simple Harmonic Motion Ques 39
- Two light springs of force constants $k _1$ and $k _2$ and a block of mass $m$ are in one line $A B$ on a smooth horizontal table such that one end of each spring is fixed on rigid supports and the other end is free as shown in the figure.
The distance $C D$ between the free ends of the spring is $60$ $cm$. If the block moves along $A B$ with a velocity $120$ $cm / s$ in between the springs, calculate the period of oscillation of the block. (Take, $\left.k _1=1.8 N / m, k _2=3.2 N / m, m=200 g\right)(1985,6 M)$
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Answer:
Correct Answer: 39.$(2.82$ $s)$
Solution:
Formula:
- Between $C$ and $D$ block will move with constant speed of $120$ $cm / s$. Therefore, period of oscillation will be (starting from $C$ ).
$ T=t _{C D}+\frac{T _2}{2}+t _{D C}+\frac{T _1}{2} $
Here, $\quad T _1=2 \pi \sqrt{\frac{m}{k _1}}$ and $T _2=2 \pi \sqrt{\frac{m}{k _2}}$
and $t _{C D}=t _{D C}=\frac{60}{120}=0.5 s$
$\therefore \quad T=0.5+\frac{2 \pi}{2} \sqrt{\frac{0.2}{3.2}}+0.5+\frac{2 \pi}{2} \sqrt{\frac{0.2}{1.8}}$
$ (m=200 g=0.2 kg) $
$ T=2.82 s $
BETA
