First slide

Simple harmonic motion

Question

Displacement-time equation of a particle executing SHM is $x = 10 \sin (\frac{π}{3} t + \frac{π}{6}) c m$ . The distance covered by particle in first 3 seconds is

Difficult

A

5 cm
B

20 cm
C

10 cm
D

15 cm

Solution

Mathematically,
Velocity of the particle is $v = 10 \frac{π}{3} \cos (\frac{π}{3} t + \frac{π}{6}) c m / s$
$\begin{array}{l} When [0 < (\frac{π t}{3} + \frac{π}{6}) < \frac{π}{2}] or [0 < t < 1 \sec], \\ v = \frac{10 π}{3} \cos (0 < θ < \frac{π}{2}) \Rightarrow v = + ve \\ When [\frac{π}{2} < (\frac{π t}{3} + \frac{π}{6}) < π] or [1 < t < 2.5 \sec] \\ v = \frac{10 π}{3} \cos (\frac{π}{2} < θ < π) \Rightarrow v = - ve \\ When [π < (\frac{π t}{3} + \frac{π}{6}) < \frac{3 π}{2}] or [2.5 < t < 4 \sec] \\ v = \frac{10 π}{3} \cos (π < θ < \frac{3 π}{2}) \Rightarrow v = - ve \end{array}$
Total distance travelled from $0 \leq t \leq 3 = \int_{0}^{3} |v| d t$
Hence total distance travelled from $0 \leq t \leq 3 \sec$
$\begin{array}{l} = \int_{0}^{3} | v | dt = \int_{0}^{1} \{\frac{10 π}{3} \cos (\frac{π}{3} t + \frac{π}{6})\} dt + \int_{1}^{3} - \{\frac{10 π}{3} \cos (\frac{π}{3} t + \frac{π}{6})\} dt \\ = {\frac{10 π}{3} \frac{3}{π} \sin (\frac{π t}{3} + \frac{π}{6})|}_{0}^{1} - {\frac{10 π}{3} \frac{3}{π} \sin (\frac{π t}{3} + \frac{π}{6})|}_{1}^{3} \\ = 10 (1 - \frac{1}{2}) - 10 (\frac{- 1}{2} - 1) \\ = 5 + 15 = 20 cm \end{array}$

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