For an SHM oscillator, the amplitude is 5 cm and its time period is 4 seconds. The minimum time taken by the particle to pass between points which are at distances 4 cm and 3 cm from the centre of oscillation on the same side of it will be

Option 1 - 0.13 second

Option 2 - 0.18 second

Option 3 - 0.26 second

Option 4 - 0.35 second

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Vishal Baghel | Contributor-Level 10

5 months ago

Correct Option - 2

Detailed Solution:

y = A sin (2πt/T)
t? - t? = (T/2π) [sin? ¹ (x? /A) - sin? ¹ (x? /A)]

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$A_{1} = \sqrt[]{{(5)}^{2} + {(5 \sqrt[]{3})}^{2}} = 10$

$A_{2} = 5$

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For an SHM oscillator, the amplitude is 5 cm and its time period is 4 seconds. The minimum time taken by the particle to pass between points which are at distances 4 cm and 3 cm from the centre of oscillation on the same side of it will be

Alok Kumar Singh

$y = A \sin ? \frac{2 π t}{T}$

$t_{2} - t_{1} = \frac{T}{2 π} [\sin^{- 1} ? \frac{x_{1}}{A} - \sin^{- 1} ? \frac{x_{2}}{A}]$

The given arrangement is in vertical plane and initially the springs are in natural length. The uniform rod is hinged at end O, about which it can freely rotate. The angular frequency of small oscillations of the rod is n/8 √(k/m). Find n. (Given: m = mass of rod, L = length of rod and k = mg/L)

Vishal Baghel

Given mg = kL
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Alok Kumar Singh

Displacement equation of SHM of frequency ' $n$ '

$x = A s i n ? (ω t) = A s i n ? (2 π n t)$

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