Motion Test - 9

According to the question, the correct option will be ‘A’ because Time translation symmetry gives rise to the conservation of energy.
Space translation symmetry gives rise to the conservation of momentum.
Rotational symmetry gives rise to the conservation of angular momentum.
The conservation law associated with this symmetry is the conservation of electrical charge.

According to the question, the correct option will be ‘B’ because

 x = sin²ωt − 2cos²ωt

= 1 − 3cos²ωt

= 1−3$\left(\frac{1+\cos 2\omega }{2}\right)$

= -$\frac{1}{2}-\frac{3}{2}\cos 2\omega t$

which is a periodic function with angular frequency of 2ω.

According to the question the correct option will be ‘B’ because

Modulation Index, \(m_{f}=\frac{\text { Maximum frequency deviation }}{\text { Modulating frequency }}\)

\(\Rightarrow \mathrm{m}_{\mathrm{f}}=\frac{10 \mathrm{kh} z}{2 \mathrm{khz}}\)

\(\Rightarrow \mathrm{m}_{\mathrm{f}}=5\)

According to Newton’s first law , body remains in its state unless acted by external force. This property of body is termed as Inertia.
Option A is correct.

According to the question the correct option will be ‘C’ because The motion is oscillatory but not periodic because the oscillations are followed from the maximum displacement (negative extreme) at t=0, the corresponding echo is equal to π/2

According to the question the correct option will be ‘B’ because time period is the time take to cover one oscillation. When a pendulum clock goes from one position to another it take 1 sec. Hence, time taken for one oscillation is 2 sec.

Given the equation of displacement of the particle, \(y=\sin ^{3} \omega t\)

We know \(\sin 3 \theta=3 \sin \theta-4 \sin ^{3} \theta\)

Hence, \(y=\frac{(3 \sin \omega t-4 \sin 3 \omega t)}{4}\)

\(\Rightarrow 4 \frac{d y}{d t}=3 \omega \cos \omega t-4 \times[3 \omega \cos 3 \omega t]\)

\(\Rightarrow 4 \times \frac{d^{2} y}{d t^{2}}=-3 \omega^{2} \sin \omega t+12 \omega \sin 3 \omega t\)

\(\Rightarrow \frac{d^{2} y}{d t^{2}}=\frac{-3 \omega^{2} \sin \omega t+12 \omega \sin 3 \omega t}{4}\)

\(\Rightarrow \frac{d^{2} y}{d t^{2}}\) is not proportional to \(y\)

Hence, the motion is not SHM. 

As the expression is involving sine function, hence it will be periodic. 

Also \(\sin ^{3} \omega t=(\sin \omega t)^{3}\)

\(=[\sin (\omega t+2 \pi)]^{3}\)

\(=[\sin (\omega t+2 \pi / \omega)]^{3}\)

Hence, \(y=\sin ^{3} \omega t\) represents a periodic motion with period \(2 \pi / \omega\)

According to the question the correct option will be ‘D’ because Oscillation is a motion in which a body moves back and forth repeatedly about a mean position. Since in all the first 3 options, the motion is oscillatory, the correct option is 'None of these'.

\(y=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t)\)

\(y=5 \times 2\left(\frac{1}{2} \sin 3 \pi t+\frac{\sqrt{3}}{2} \cos 3 \pi t\right)\)

\(y=10\left(\cos \frac{\pi}{3} \sin 3 \pi t-\sin \frac{\pi}{3} \cos 3 \pi t\right)\)

\(y=10 \sin \left(3 \pi t+\frac{\pi}{3}\right)\)

Comparing it with \(y=A \sin (\omega t+\phi)\)

\(\Rightarrow A=10\) i.e., amplitude is \(10 \mathrm{cm}\)

\(\Rightarrow T=\frac{2 \pi}{\omega}=\frac{2 \pi}{3 \pi}=\frac{2}{3}\) sec. i.e, time period is \(\frac{2}{3}\) sec.

According to the question the correct option will be ‘C’ because Inertia is the tendency of an object to remain at rest or in motion. Newton's First Law of Motion states that an object will remain at rest or move at a constant speed in a straight line unless it is acted on by an unbalanced force.

So, when a body is at rest, it is obeying Newton's first law or motion or law of inertia.

According to the question the correct option will be ‘D’ because Firstly, given that it is a thin shell i.e. hollow sphere then it will perform oscillation on SHM.

In case of solid sphere it performs SHM.

Frictional force is a constant force acting opposite to the direction of motion

A constant force only makes an object to move in one direction

A time varying force makes an object to move in one direction with its speed increasing

Force due to spring, keeps varying, as the object is moved away from its mean position and thus acts as a restoring force

The correct option is (d)

According to the question the correct option will be ‘B’ because Oscillatory motion is the to and fro motion of a particle which is repeated over a fixed interval of time.
Clock does not show oscillatory motion. It shows periodic motion not oscillatory motion.
Clock does not show to and fro position about a mean position.

According to the question the correct option will be ‘D’ because 

K.E=$\frac{1}{2}$mω²(A²−x²)

(K.E)_max=$\frac{1}{2}$mω²A²

(K.E)_min=0

P.E = $\frac{1}{2}$mω²x²

(P.E)_max=$\frac{1}{2}$mω²A²

According to the question the correct option will be ‘B’ because Simple harmonic motion provides a basis for the characterization of more complicated motions through the techniques of Fourier analysis. Here any waveform can be represented as closely as desired by the combination of a sufficiently large number of sinusoidal waves that form a harmonic series.   Fourier's theorem suggests that any periodic function can be represented as an algebraic sum of sine and cosine functions called a Fourier series.