A point performs simple harmonic oscillation of period T and the equation of motion is given by x = asin(ωt + π/6). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
A point performs simple harmonic oscillation of period \(T\) and the equation of motion is given by \(x = a \sin(\omega t + \pi/6)\). After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
(a) \( T/3 \)
(b) \( T/12 \)
(c) \( T/8 \)
(d) \( T/6 \)
Detailed Answer:
The equation of motion for the particle is:
\[
x = a \sin(\omega t + \pi/6),
\]
where \(\omega = \frac{2\pi}{T}\). The velocity \(v\) is given by:
\[
v = \frac{dx}{dt} = a \omega \cos(\omega t + \pi/6).
\]
The maximum velocity \(v_{\text{max}}\) occurs when \(\cos(\omega t + \pi/6) = 1\):
\[
v_{\text{max}} = a \omega.
\]
Now, the velocity is half of the maximum velocity:
\[
v = \frac{v_{\text{max}}}{2} = \frac{a \omega}{2}.
\]
Substituting into the velocity equation:
\[
\frac{a \omega}{2} = a \omega \cos(\omega t + \pi/6).
\]
Simplify by dividing through by \(a \omega\):
\[
\frac{1}{2} = \cos(\omega t + \pi/6).
\]
Solve for \(\omega t + \pi/6\):
\[
\cos^{-1}\left(\frac{1}{2}\right) = \omega t + \pi/6.
\]
The value of \(\cos^{-1}(1/2)\) is \(\pi/3\). Therefore:
\[
\omega t + \pi/6 = \pi/3.
\]
Subtract \(\pi/6\):
\[
\omega t = \pi/3 - \pi/6 = \pi/6.
\]
Substitute \(\omega = \frac{2\pi}{T}\):
\[
\frac{2\pi}{T} t = \frac{\pi}{6}.
\]
Solve for \(t\):
\[
t = \frac{T}{12}.
\]
Hence, the correct answer is:
\[
\boxed{T/12.}
\]
.png)
. After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?){kind=link}
0 Comments