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PART - 2 CH - 13 Oscillations — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 13 Oscillations5 Marks
Question
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial $(t=0)$ position of the particle, the readius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case ( $x$ is in cm and $t$ is in s).
(a) $x=-2 \sin (3 t+\pi / 3)$
(b) $x=\cos (\pi / 6-t)$
(c) $x=3 \sin (2 \pi t+\pi / 4)$
(d) $x=2 \cos \pi t$
✓

Answer

(a) From given equation
$x=-2 \sin \left(3 t +\frac{\pi}{3}\right)$
$2 \sin \left(3 t+\pi+\frac{\pi}{3}\right)$$\ldots$(1)
Equation (1) by comparing $x= A \sin (\omega t +\phi)$
$
\phi=\pi+\frac{\pi}{3}=\frac{4 \pi}{3}, A=2 cm . \omega=3 radian / sec .
$
Image
(b) $x=\cos \left(\frac{\pi}{6}- t \right)=\cos \left( t -\frac{\pi}{6}\right)$
$[\because \cos (-\theta)=\cos \theta]$
$x=1 \cos \left(t-\frac{\pi}{6}\right)$$\ldots$(2)
The corresponding reference circle is shown in the figure, comparing equation (2) with the following equation :
$x= A \cos (\omega t +\phi)$$\ldots$(3)
A = 1 cm
$\omega=1 radian / sec . T =2 \pi sec$. and $\phi=-\frac{\pi}{6}$.
Hence the given equation can be expressed through the following circle diagram :
Image
(c) $x=3 \sin \left(2 \pi t +\frac{\pi}{4}\right) cm$.
Comparing equation (4) with $x= A \sin (\omega t +\phi)$
Amplitude $A =3 cm, \omega=2 \pi$
$
T=\frac{2 \pi}{\omega}=1 sec . \phi=-\frac{\pi}{4}
$
The corresponding reference circle is shown in the figure below :
Image
(d) $x=2 \cos \pi t$
$
=2 \cos (\pi t+0) cm .
$
The corresponding reference circle is shown in the figure comparing equation (5) $x= A \cos \omega t$
$
A=2 cm, \omega=\pi, T=2, \phi=0
$
Image

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