Question

A body oscillates with Simple Harmonic Motion according to the equation $x=5.0\cos (2\pi t+\pi )$. At time $t=1.5s$ , its displacement, speed and acceleration respectively are:
(A). \[0,-10\pi ,20{{\pi }^{2}}\]
(B). \[5,0,-20{{\pi }^{2}}\]
(C ). \[2.5,+20\pi ,0\]
(D). \[-5.0,+5\pi ,-10{{\pi }^{2}}\]

Answer 1

Hint: If the equation of SHM is given, then we can easily calculate its speed and acceleration. Speed is a derivative of displacement with respect to time and acceleration can be calculated by taking the double derivative of displacement with respect to time.

Complete step-by-step solution -

Firstly, we will calculate the displacement of the body undergoing Simple harmonic motion at a time t= 1.5sec,
According to the question:
$x=5.0\cos (2\pi t+\pi )$
This is our equation (1)
Substituting the value of t
$t=1.5$
$\Rightarrow x=5.0\cos (2\pi \times t+\pi )$
$\Rightarrow x=5.0\cos (2\pi \times 1.5+\pi )$
$\Rightarrow x=5.0\cos (3\pi +\pi )$
$\Rightarrow x=5.0\cos (4\pi )$

We know,
\[\cos \left( n\pi \right)={{\left( -1 \right)}^{n}}\]
\[\Rightarrow x=5\]
Hence, displacement is \[x=5\]
Now, to calculate speed of the body we need to differentiate equation (1) with respect to time
On differentiating we get
$\Rightarrow \dfrac{dx}{d{{t}^{{}}}}=\dfrac{d\left[ 5.0\cos \left( 2\pi t+\pi \right) \right]}{d{{t}^{{}}}}$
$\Rightarrow \dfrac{dx}{d{{t}^{{}}}}=-5.0\times 2\pi\sin \left( 2\pi t+\pi \right)$
$\Rightarrow \dfrac{dx}{d{{t}^{{}}}}=-10.0\times \pi\sin \left( 2\pi \times 1.5+\pi \right)$
$\Rightarrow \dfrac{dx}{d{{t}^{{}}}}=-10.0\times \pi\sin \left( 4\pi \right)$
We know,
$\sin \left( n\pi \right)=0$
$\Rightarrow \dfrac{dx}{dt}=0$
Therefore, the speed of the oscillating body is 0
To calculate acceleration, we will double differentiate equation (1) with respect to time.
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=\dfrac{{{d}^{2}}\left[ 5.0\cos \left( 2\pi t+\pi \right) \right]}{d{{t}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-20.0{{\pi }^{2}}\cos \left( 2\pi t+\pi \right)$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-20.0{{\pi }^{2}}\cos \left( 2\pi \times 1.5+\pi \right)$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-20.0{{\pi }^{2}}\cos \left( 3\pi +\pi \right)$
$\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-20.0{{\pi }^{2}}\cos \left( 4\pi \right)$

We know,
\[\cos \left( n\pi \right)={{\left( -1 \right)}^{n}}\]
\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-20.0{{\pi }^{2}}\]
Hence the acceleration of the oscillating body is $-20{{\pi }^{2}}$
The correct answer is B. \[5,0,-20{{\pi }^{2}}\]

Additional information:
Simple harmonic motion is a periodic motion in which a body oscillates about a mean equilibrium position under a restoring force which is directly proportional to the displacement of the body from the mean position. The formula for SHM is given by:
$x={{x}_{0}}\cos (\omega t+\phi )$

Note: Students should remember that acceleration and displacement are in opposite directions that is why in the answer too acceleration is negative while displacement is positive. Also, to calculate instantaneous time the value of time should be substituted directly in place of ‘t’ in the given equation.