Physics M.C.Q (1 Marks)

1. A

Explanation:

We know the resultant amplitude Is given by

$\text{R}_\text{net}=\sqrt{\text{A}^2+\text{A}^2+2\text{A}^2\cos120^\circ}$ $(\phi=120^\circ)$

$=\sqrt{2\text{A}^2-\text{A}^2}$ $\Big[\because\cos120^\circ=\frac{-1}{2}\Big]$

$=\text{A}$

1. Inverse of its length.

Explanation:

The relation between wave speed and the length of the string is given by

$\text{v}=\frac{1}{2\text{l}}\sqrt{\frac{\text{F}}{\mu}}$

where

l is the length of the string

F is the tension

$\mu$ linear mass density

From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.

$\text{v}\propto\frac{1}{\text{l}}.$

1. $\text{v}$

Explanation:

The boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.

2. Amplitude $\frac{\text{A}}{2},$ frequency $\frac{\omega}{\pi}$

Explanation:

$\text{y}=\text{A}\sin^2(\text{kx}-\omega\text{t})$

$\Big[\cos^2\theta=1-2\sin^2\theta,\ \sin^2\theta=\frac{1-\cos^2\theta}{2}\Big]$

$\text{y}=\text{A}\Big[\frac{1-\cos^2(\text{kx}-\omega\text{t})}{2}\Big]$

$\text{y}=\frac{\text{A}}{2}\Big[1-\cos^2(\text{kx}-\omega\text{t})\Big]$

Thus, we have:

Amplitude $=\frac{\text{A}}{2}$

Frequency $=2\Big(\frac{\omega}{2\pi}\Big)$

$=\frac{\omega}{\pi}$

2. Vibrate with a frequency of 208Hz.

Explanation:

According to the relation of the fundamental frequency of a string

$\nu=\frac{1}{2\text{l}}\sqrt{\frac{\text{F}}{\mu}}$

Where

l Is the length of the string

F Is the tension

$\mu$ Is the linear mass density

We know that v₁ = 416Hz, l₁ = l and l₂ = 2l.

$\text{v}_1\propto\frac{1}{\text{l}_1}$

$\text{v}_1\text{l}_1=\text{v}_2\text{l}_2$

$(416)\text{l}=\text{v}_2(2\text{l})$

$\text{v}_2=208\text{Hz}$

4. Between 0 and 2A

Explanation:

The amplitude of the resultant wave depends on the way two waves superimpose, i.e., the phase angle $\phi.$ So, the resultant amplitude lies between the maximum resultant amplitude (A_max) and the minimum resultant amplitude (A_min).

A_max = A + A = 2A

A_min = A - A = 0

4. 16kg.

Explanation:

According to the relation of the fundamental frequency of a string

$\nu=\frac{1}{2\text{l}}\sqrt{\frac{\text{F}}{\mu}}$

Where l is the length of the string

F is the tension

$\mu$ is the linear mass density of the string

We know that v₁ = 416Hz, l₁ = l and l₂ = 2l

Also, m₁ = 4kg and m₂ = ?

$\nu_1=\frac{1}{2\text{l}_1}\sqrt{\frac{\text{m}_1\text{g}}{\mu}}\ \dots(1)$

$\nu_2=\frac{1}{2\text{l}_2}\sqrt{\frac{\text{m}_2\text{g}}{\mu}}\ \dots(2)$

So, in order to maintain the same fundamental mode

$\nu_1=\nu_2$

squaring both sides of equations (1) and (2) and then equating

$\frac{1}{4\text{l}^2}\frac{4\text{g}}{\mu}=\frac{1}{16\text{l}^2}\frac{\text{m}_2\text{g}}{\mu}$

$\Rightarrow\text{m}_2=16\text{kg}$

2. 2A₂

Explanation:

we know resultant amplitude is given by

$\text{A}_\text{net}=\sqrt{\text{A}_1^2+\text{A}_2^2+2\text{A}_1\text{A}_2\cos\phi}$

For maximum resultant amplitude

$\text{A}_\text{max}=\text{A}_1+\text{A}_2$

For mnimum resultant amplitude

$\text{A}_\text{min}=\text{A}_1-\text{A}_2$

So, the difference between A_max and A_min is

$\text{A}_\text{max}-\text{A}_\text{min}=\text{A}_1+\text{A}_2-\text{A}_1+\text{A}_2=2\text{A}_2$

3. $\lambda'<\lambda$

Explanation:

As $\text{v}=\sqrt{\frac{\text{f}}{\mu}}$

A wave pulse travels faster in a thinner string. The wavelength of the transmitted wave is equal to the wavelength of the Incident wave because the frequency remains constant.

3. The frequency $=\frac{25}{\pi}\text{Hz}$

4. The amplitude = 0.001mm.

Explanation:

$\text{y}=(0.001\text{mm})\sin\Big[(50\text{s}^{-1})\text{t}+(2.0\text{m}^{-1})\text{x}\Big]$

Equating the above equation with the general equation, we get:

$\text{y}=\text{A}\sin(\omega\text{t}-\text{kx})$

$\omega=\frac{2\pi}{\text{T}}=2\pi\nu$

$\text{k}=\frac{2\pi}{\lambda}$

Here. A is the amplitude, $\omega$ is the angular frequency, k is the wave number and $\lambda$ is the wavelength.

$\text{A}=0.001\text{mm}$

Now,

$50=2\pi\nu$

$\Rightarrow\nu=\frac{25}{\pi}\text{Hz}$

2. A and B move in opposite directions.

4. The displacements at A and B have equal magnitudes.

Explanation:

A and Bhave a phase difference of fl. So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by any odd multiple of their wavelength.

$\overrightarrow{\text{yA}}=\text{A}\sin(\omega\text{t})$

$\overrightarrow{\text{yB}}=\text{B}\sin(\omega\text{t}+\pi)$

3. $\frac{\lambda}{2}$

Explanation:

A sine wave has a maxima and a minima and the particle displacement has phase difference of $\pi$ radians. Therefore, applying similar argument we can say that if a particular particle has zero displacement at a certain instant. then the particle closest to it having zero displacement is at a distance is equal to $\frac{\lambda}{2}.$

3. $\frac{\lambda}{2}$

 Explanation:

A sme wave has a maxima and a minimum and the particle displacement has phase difference of n radians. The speeds at the maximum point and at the minimum point are same although the direction of motion are different. The difference between the posiuons of maxima and minima Is equal to $\frac{\lambda}{2}.$

1. $\frac{1}{2}$

Explanation:

wave speed is given by $\nu=\sqrt{\frac{\text{T}}{\mu}}$

Where

T is the tension in the string

v is the speed of the wave

$\mu$ Is the mass per unit length of the stnng

$\mu=\frac{\text{M}}{\text{L}}=\rho\frac{\text{V}}{\text{L}}=\rho\frac{(\text{AL})}{\text{L}}$

Where

Mis the mass of the stnng. which can be written as pV

L Is the length of the string

$=\rho(\pi\text{r}^2)=\rho\Big(\pi\frac{\text{D}^2}{4}\Big)$

$\therefore\nu=\sqrt{\frac{\text{T}}{\rho\pi\frac{\text{D}^2}{4}}}=\frac{2}{\text{D}}\sqrt{\frac{\text{T}}{\rho\pi}}$

Where Dis the diameter of the string.

Thus, $\text{V}\propto\frac{1}{\text{D}}$

Since, $\text{r}_\text{A}=2\text{r}_\text{B}$

$\text{v}_\text{A}\propto\frac{1}{2\text{r}_\text{A}}\propto\frac{1}{2\times2\text{r}_\text{B}}\ \dots(1)$

$\text{v}_\text{B}\propto\frac{1}{2\text{r}_\text{B}}\ \dots(2)$

From Equations (1) and (2) we get

$\frac{\text{v}_\text{A}}{\text{v}_\text{B}}=\frac{1}{2}$

1. Must be an integral multiple of $\frac{\lambda}{4}$

Explanation:

A standing wave is produced on a string clamped at one end and free at the other. Its fundamental frequency is given by

$\nu=\Big(\text{n}+\frac{1}{2}\Big)\frac{\nu}{2\text{L}}$

$\Rightarrow\text{v}=\nu\lambda$

$\Rightarrow\nu=\Big(\text{n}+\frac{1}{2}\Big)\frac{\nu\lambda}{2\text{L}}$

$\Rightarrow\text{L}=\Big(\frac{2\text{n}+1}{4}\Big)\lambda$

$\Rightarrow\text{L}=\frac{\lambda}{4},\frac{3\lambda}{4},\ \dots$

1. $\text{x}=\text{A}\sin(\text{ky}-\omega\text{t})$

Explanation:

Here x is the particle displacement of the wave and the wave is travelling along the Y-axis because the particle displacement is perpendicular to the direction of wave motion.

2. $\sqrt{2}\text{a}$

Explanation:

We know that the resultant of the amplitude is given by

$\text{R}_\text{net}=\sqrt{\text{A}_1^2+\text{A}_2^2+2\text{A}_1\text{A}_2\cos\phi}$

For the particular case, we can write

$=\sqrt{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}$

$=\sqrt{2}\text{a}$

4. The information is insufficient to find the relation betwee t₁ and t₂.

Explanation:

$\text{v}=\sqrt{\frac{\eta}{\rho}}$

But because the length of wires A and B is not known, the relation between A and B cannot be determined.

4. $\frac{1}{\sqrt{2}}$

Explanation:

T_AB = T

T_CD = 2T

where

T_AB is the tension in the string AB

T_CD is the tension in the stnng CD

The eelatlon between tension and the wave speed is given by

$\text{v}=\sqrt{\frac{\text{T}}{\mu}}$

$\text{v}\propto\sqrt{\text{T}}$

where

v is the wave speed of the transverse wave

$\mu$ is the mass per unit length of the string

$\frac{\text{v}_1}{\text{v}_2}=\sqrt{\frac{\text{T}}{2\text{T}}}$

$=\frac{1}{\sqrt{2}}$