Question

A particle performs linear SHM of period 12 seconds and amplitude 8 cm. How long will the particle take to travel 6 cm from the extreme position?
A) 3.517 s
B) 2.517 s
C) 20.517 s
D) 8.517 s

Answer 1

Hint: The displacement of the particle can be given by the difference between the initial and final positions and as the particle is performing SHM, the expression for its displacement will be:
$x = A\cos \omega t$
and
$\omega = \dfrac{{2\pi }}{t}$ rad/sec
Using these, the time of required travel can be calculated

Complete step by step answer:
Given:
Time period (t) = 12 s
Amplitude (A) = 8 cm
Displacement (x) of the particle will be equal to the difference between the amplitude (A) and the distance to be travelled from the mean position:
x = (8 – 6) cm
x = 2 cm
Since, the particle is in simple harmonic motion its displacement from the positive extreme position is given as:
$x = A\cos \omega t$ ____________ (1)
We know,
$\omega = \dfrac{{2\pi }}{t}$ [As t = 12 s]
$\omega = \dfrac{\pi }{6}rad/s$
Substituting all the given values in (1), we get:
\[
  x = A\cos \omega t \\
  2 = 8\cos \left( {\dfrac{\pi }{6}t} \right) \\
  \cos \left( {\dfrac{\pi }{6}t} \right) = 0.25 \\
  \dfrac{\pi }{6}t = {\cos ^{ - 1}}0.25 \\
  {\cos ^{ - 1}}0.25 \approx {75.52^ \circ } \\
 \]
Converting radians into degree:
$75.52 \times \dfrac{\pi }{{180}}rad$
Now, calculating for t:
\[
  \dfrac{\pi }{6}t = 75.52 \times \dfrac{\pi }{{180}} \\
  t = \dfrac{{6 \times 75.52}}{{180}} \\
  t = \dfrac{{75.52}}{{30}} \\
  t = 2.517s \\
    \\
 \]
Therefore, the particle performing linear SHM will take 2.517 seconds to travel 6 cm from the extreme position and thus the correct option is B).

Note:Displacement refers to the difference between the final and initial positions of an object.
SHM is linear because it is always in a straight line and on a graph, it can be denoted as: