The periodic time of a particle executing SHM is 4 s. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

Question

The periodic time of a particle executing SHM is 4 s. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is (A)  1s  (B)  2s  (C)  (1/3)s  (D)  (2/3)s

Tardigrade Admin · Accepted Answer

y=A sin ( (2π /T) )t ⇒ (A/2)=A sin ( (2π /4) )t ⇒ (π t/2)=(π /6)⇒ t=(1/3)s

Q. The periodic time of a particle executing SHM is 4 s. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

$1 s$

$2 s$

$\frac{1}{3} s$

$\frac{2}{3} s$

$y = A sin (\frac{2 π}{T}) t$ $\Rightarrow$ $\frac{A}{2} = A sin (\frac{2 π}{4}) t$ $\Rightarrow$ $\frac{π t}{2} = \frac{π}{6} \Rightarrow t = \frac{1}{3} s$

Q. The periodic time of a particle executing SHM is 4 s. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

1s

2s

31​s

32​s

y=Asin(T2π​)t ⇒ 2A​=Asin(42π​)t ⇒ 2πt​=6π​⇒t=31​s

$1 s$

$2 s$

$\frac{1}{3} s$

$\frac{2}{3} s$

$y = A sin (\frac{2 π}{T}) t$ $\Rightarrow$ $\frac{A}{2} = A sin (\frac{2 π}{4}) t$ $\Rightarrow$ $\frac{π t}{2} = \frac{π}{6} \Rightarrow t = \frac{1}{3} s$