Physics M.C.Q (1 Marks)

4. $\text{y = a}\tan\omega\text{t}$

Explanation:

S.H.M. is one which is bounded within well defined limits, periodic and oscillatory. The first four equations represent S.H.M. but the fifth equation does not represent S.H.M.

3. $\pm\text{a}$

Explanation:

In S.H.M., the maximum P.E. is at the extreme position and maximum K.E. is at the mean position.

1. Proportional to $\frac{1}{\sqrt{\text{a}}}$

Explanation:

Potential energy, $\text{U = kx}^3$

Force, $\text{F}=\frac{-\text{dU}}{\text{dx}}=-3\text{kx}^2$

max. force $=\text{F}_{\text{max}}-3\text{kx}^2=-\text{m}\omega^2\text{a}$

$\omega^2=\frac{3\text{ka}^2}{\text{ma}}=\frac{3\text{ka}}{\text{m}}$

$\frac{4\pi^2}{\text{T}^2}=\frac{3\text{ka}}{\text{m}}$

$\text{T}^2\propto\frac{1}{\text{a}}$

$\text{T}\propto\frac{1}{\sqrt{\text{a}}}$

1. T₁ < T₂

Explanation:

$\frac{\text{A}}{2}=\text{A}\sin\omega\text{T}_1$

$\sin\omega\text{T}_1=\frac{1}2{}=\sin\frac{\pi}{6}$

$\text{T}_1=\frac{\pi}{6\omega}$

$\sin\omega(\text{T}_1+\text{T}_2)=1=\sin\frac{\pi}{2}$

$\omega(\text{T}_1+\text{T}_2)=\frac{\pi}{2}$

$\text{T}_1+\text{T}_2=\frac{\pi}{2\omega}$

$\text{T}_2=\frac{\pi}{2\omega}-\frac{\pi}{6\omega}=\frac{2\pi}{6\omega}=2\text{T}_1$

So, $\text{T}_1<\text{T}_2$

3. $\omega$

3. $\text{T}\propto\frac{1}{\sqrt{\text{A}}}$​​​​​​​

3. Simple harmonic with period $\frac{2\pi}{\omega}.$

Explanation:

All sine and cosine functions of t are simple harmonic in nature.

Hence the motion is simple harmonic motion.

A simple harmonic motion is always periodic.

$\text{Time period}=\text{T}'=\frac{2\pi}{\omega'}$

hence the motion is simple harmonic with time period $\frac{2\pi}{\omega}.$

4. $\frac{9}{4}$

1. 1Hz

Explanation:

Max speed $\text{v}_{\text{m}}=\text{r}\omega=\text{r}.2\pi\text{v}$

Max speed $\text{v}=\frac{\text{v}_{\text{m}}}{2\pi\text{r}}=\frac{31.4}{2\times3.14\times5}=1\text{Hz}$

3. Simple harmonic and time period is independent of the density of the liquid.

Explanation:

Restoring Force F = Weight of liquid column of height 2y

$\Rightarrow\text{FF}=-(\text{A}\times2\text{y}\times\rho)\times\text{g}=-2\text{A}\rho\text{gy}$

$\Rightarrow\text{F}\propto-\text{y}$

Motion is SHM with force constant

$\text{K}=2\text{A}\rho\text{g}$

$\Rightarrow\text{Time period}\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}}}$

$=2\pi\sqrt{\frac{\text{A}\times2\text{h}\times\rho}{2\text{A}\rho\text{g}}}=2\pi\sqrt{\frac{\text{h}}{\text{g}}}$

Which is independent of the density of the liquid.

3. $\frac{1}{2}\text{s}$

2. Phase of the oscillator is same at t = 2 s and t = 6 s.

4. Phase of the oscillator is same at t = 1 s and t = 5 s.

Explanation:

Two particles are said to be in same phases if the mode of vibration is same i.e, their distance will be $\text{n}\lambda(/\text{n}=1,2,3...)$

Distance between particles at t = 0 and t = 2 is $\frac{\lambda}{2}.$

So, articles are not in same phase.

As from figure the particles at t = 2 sec are at distance $\lambda$, so are in same phase.

Particles at t = 1, t = 7 are the distance $\lambda+\frac{\lambda}{2}=\frac{3\lambda}{2}$ so are not in phase.

Particles at t = 1 and 5sec are at distance = $\lambda$ so are in same phase.

4. $\text{x}=\text{A}\sin\omega\text{t}\cos^2\omega\text{t}$

1. Force acting is directly proportional to displacement from the mean position and opposite to it.

2. Motion is periodic.

4. The velocity is periodic.

Explanation:

Let us write the equation for the SHM $\text{x}=\text{a}\sin(\omega\text{t}+\phi)$

Clearly, it is a periodic motion as it involves sine function.

Let us find velocity of the particle, $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin(\omega\text{t}+6\phi))$

Velocity is also periodic because it is a cosine function.

Now let us find acceleration, $\text{A}=\frac{\text{dv}}{\text{dt}}=\frac{\text{d}^2\text{x}}{\text{dt}^2}=-\text{a}\omega^2\sin(\omega\text{t}+\phi)$

The acceleration is a sine function, hence cannot be constant.

$\Rightarrow\text{A}=-(\omega^2\text{a})\sin(\omega\text{t}+\phi)=-\omega^2\text{x}$

Force, F = mass × acceleration.

$=\text{mA}=-\text{m}\omega^2\text{x}$

Hence, force acting is directly proportional to displacement from the mean position and opposite to it.

3. Simple harmonic and time period is independent of the density of the liquid.

Explanation:

As $\text{T}=2\pi\sqrt{\frac{\text{l}}{\text{g}}}$ (where l is the length ofthe oscillating liquid column in each limb), it is independent of the density of the liquid.

3. $\sqrt{2\text{gl}(1-\cos\theta)}$

Explanation:

The height of fall of pendulum, $\text{h = l}(1-\cos\theta)$ and $\text{v}=\sqrt{2\text{gh}}$

1. $\sqrt{\text{v}_1^2+\text{v}_2^2}.$

Explanation:

When the mass is connected to the two springs individually.

$\text{v}_1=\frac{1}{2\pi}\sqrt{\frac{\text{k}_1}{\text{m}}}\ ...(1)$

$\text{v}_2=\frac{1}{2\pi}\sqrt{\frac{\text{k}_2}{\text{m}}}\ ...(2)$

Now, the block is connected with two springs considered as parallel.

Here equivalent spring constant $\text{K}_\text{eq}=\text{K}_1+\text{K}_2$

Time period of oscillation of the spring block-system is

$\text{T}=2\pi\sqrt{\frac{\text{m}}{\text{k}_\text{eq}}}=2\pi\sqrt{\frac{\text{m}}{\text{k}_1+\text{k}_2}}$

Hence frequency,

$\text{v}=\frac{1}{\text{T}}=\frac{1}{2\pi}\times\sqrt{\frac{\text{k}_1+\text{k}_2}{\text{m}}}\ ...(3)$

$\text{v}=\frac{1}{2\pi}\Big[\frac{\text{k}_1}{\text{m}}+\frac{\text{k}_2}{\text{m}}\Big]^{\frac{1}{2}}$

From Eq. (i) $\frac{\text{k}_1}{\text{m}}=4\pi^2\text{v}_1^2$ and from Eq.(ii) $\frac{\text{k}_2}{\text{m}}=4\pi^2\text{v}_2^2$

$\Rightarrow\text{v}=\frac{1}{2\pi}\Big[\frac{4\pi^2\text{v}_1^2}{1}+\frac{4\pi^2\text{v}_2^2}{1}\Big]^\frac{1}{2}=\frac{2\pi}{2\pi}[\text{v}_1^2+\text{v}_2^2]^\frac{1}{2}$

$\Rightarrow\text{v}=\sqrt{\text{v}_1^2+\text{v}_2^2}$

1. Oscillatory but not periodic.

Explanation:

$\text{x}=\text{a}\cos(\propto\text{t})^2$ is a cosine function and x varies between -a and +a, the motion is oscillatory. Now checking for periodic motion, putting t + T in place of t. T is supposed as period of the function ω(t).

$\text{x}(\text{t}+\text{T})=\text{a}\cos[\alpha(\text{t}+\text{T})]^2$

$=\text{a}\cos[\alpha^2\text{t}^2+\text{a}^2\text{T}^2+2\alpha^2\text{tT}]\neq\text{x}(\text{t})$

Hence, it is not periodic.

3. $\frac{16}{9}$

Explanation:

$\text{T}=2\pi\sqrt{\frac{\text{M}}{\text{k}}}$ and $\text{T}'=2\pi\sqrt{\frac{\text{M + m}}{\text{k}}}$

$\frac{\text{T}'}{\text{T}}=\Big(\frac{\text{M + m}}{\text{M}}\Big)^{\frac{1}{2}}=\Big(1+\frac{\text{m}}{\text{M}}\Big)^{\frac{1}2{}}$

$\frac{\text{m}}{\text{M}}=\Big(\frac{\text{T}'}{\text{T}}\Big)^2-1=\Big(\frac{5}{3}\Big)^2-1=\frac{16}{9}$

3. A circle.

Explanation:

Resultant displacement is x + y

$\text{x}=\text{a}\cos\omega\text{t}\ ...(1)$

$\text{y}=\text{A}\sin\omega\text{t}\ ...(2)$

Dispacement $=\text{a}\cos\omega\text{t}+\text{a}\sin\omega\text{t}$

$\text{y}'=\text{a}\sqrt{2}\Big[\frac{\cos\omega\text{t}}{\sqrt{2}}+\frac{\sin\omega\text{t}}{\sqrt{2}}\Big]$

$\text{y}'\text{a}\sqrt{2}\ [\cos\omega\text{t}\cos45^\circ+\sin\omega\text{t}\sin45^\circ]$as the particle is acted simultaneously by Mutually perpendicular direction.

$\text{y}'=\text{a}\sqrt{2}\cos\text{s}(\omega\text{t}-45^\circ)$

Hence the displacement is neither a straight line nor a parabola.

Now, squaring and adding (i), (ii)

$\text{x}^2+\text{y}^2=\text{a}^2\cos^2\omega\text{t}+\text{a}^2\sin^2\omega\text{t}=\text{a}^2[\cos^2\omega\text{t}+\sin^2\omega\text{t}]$

$\text{x}^2+\text{y}^2=\text{a}^2$

This shows the equation of circle, hence the motion is circular motion.

3. Oscillatory but not periodic.

Explanation:

Since x varies between -a and +a, the motion is oscillatory.

But as $\cos(\alpha\text{t}^2)=\cos(\alpha^2\text{t}^2)$ does not have a period, the motion is not periodic.

2. $0$

Exlanation:

Velocity of the particle executing SHM is given as $\text{v}=\omega\sqrt{\text{A}^2-\text{x}^2}$

At extreme position, x = A

⇒ v = 0