A particle performs simple harmonic motion from mean position, with period $8 \mathrm{~s}$. The distance travelled by it between 1 st and 2 nd second of its motion is (A = amplitude of S. H. M.)A. $A \frac{1}{2}$B. $A\left(\frac{1}{\sqrt{2}}-1\right)$C. $A\left(1-\frac{1}{\sqrt{2}}\right)$D. $A\left(\frac{1}{\sqrt{2}}\right)$

Home » A particle performs simple harmonic motion from mean position, with period . The distance travelled by it between 1 st and 2 nd second of its motion is (A = amplitude of S. H. M.)A. B. C. D.

A particle performs simple harmonic motion from mean position, with period $8 \mathrm{~s}$ . The distance travelled by it between 1 st and 2 nd second of its motion is (A = amplitude of S. H. M.)
A. $A \frac{1}{2}$
B. $A\left(\frac{1}{\sqrt{2}}-1\right)$
C. $A\left(1-\frac{1}{\sqrt{2}}\right)$
D. $A\left(\frac{1}{\sqrt{2}}\right)$

Physics

A particle performs simple harmonic motion from mean position, with period $8 \mathrm{~s}$ . The distance travelled by it between 1 st and 2 nd second of its motion is (A = amplitude of S. H. M.)
A. $A \frac{1}{2}$
B. $A\left(\frac{1}{\sqrt{2}}-1\right)$
C. $A\left(1-\frac{1}{\sqrt{2}}\right)$
D. $A\left(\frac{1}{\sqrt{2}}\right)$

Correct answer is C.

In $2 \mathrm{~s}$ (which is equal to $\frac{T}{4}$ ), one amplitude will be convered. In $1 \mathrm{st}$ second $x=a \sin \left(\frac{\pi}{4}\right)=\frac{a}{\sqrt{2}}$
Required ratio $=\frac{\frac{a}{\sqrt{2}}}{a-\frac{a}{\sqrt{2}}}=\frac{1}{\sqrt{2}-1}$

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