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Q.

A particle executes SHM of type $x = a \sin ω t$ . It takes time $t_{1}$ from $x = 0$ to $x = \frac{a}{2}$ and $t_{2}$ from $x = \frac{a}{2}$ to $x = a$ . The ratio of $t_{1} : t_{2}$ will be :

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a

$1 : 3$

b

$2 : 1$

c

$1 : 2$

d

$1 : 1$

answer is B.

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Detailed Solution

Time required between mean to half of amplitude $(\frac{a}{2})$ position in $t_{1}$ is $= \frac{T}{12}$

Time required between $(\frac{a}{2})$ to a position is $t_{2}$ = $\frac{T}{6}$

$\frac{t_{1}}{t_{2}} = \frac{\frac{T}{12}}{\frac{T}{6}} = \frac{1}{2} = 1 : 2$

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A particle executes SHM of type x = asinωt. It takes time t1 from x = 0 to x = a2 and t2 from x = a2 to x = a. The ratio of t1:t2 will be :