Neet Solved Paper 2015 Question 24
Question: A particle is executing SHM along a straight line. Its velocities at distances $ x _1 $ and $ x _2 $ from the mean position are $ v _1 $ and $ v _2 $ , respectively. Its time period is
Options:
A) $ 2\pi \sqrt{\frac{x_1^{2}+x_2^{2}}{v_1^{2}+v_2^{2}}} $
B) $ 2\pi \sqrt{\frac{x_2^{2}-x_1^{2}}{v_1^{2}-v_2^{2}}} $
C) $ 2\pi \sqrt{\frac{v_1^{2}+v_2^{2}}{x_1^{2}+x_2^{2}}} $
D) $ 2\pi \sqrt{\frac{v_1^{2}-v_2^{2}}{x_1^{2}-x_2^{2}}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Let A be the amplitude of oscillation then $ v_1^{2}={{\omega }^{2}}(A^{2}-x_1^{2}) $ …(i)
$ v_2^{2}={{\omega }^{2}}(A^{2}-x^{2})\ $ …(ii)
Subtracting Eq. (ii) from Eq. (i),
we get $ v_1^{2}-v_2^{2}={{\omega }^{2}}(x_2^{2}-x_1^{2}) $
$ \Rightarrow \omega =\sqrt{\frac{v_1^{2}-v_2^{2}}{x_2^{2}-x_1^{2}}} $
$ \Rightarrow \frac{2\pi }{T}=\sqrt{\frac{v_1^{2}-v_2^{2}}{x_2^{2}-x_1^{2}}} $
$ \Rightarrow \ T=2\pi \sqrt{\frac{x_2^{2}-x_1^{2}}{v_1^{2}-v_2^{2}}} $
