Physics M.C.Q (1 Marks)

1. Every particle has a fixed amplitude which is different from the amplitude of its nearest particle

2. All the particles cross their mean position at the same time.  

4.  There is no net transfer of energy across any plane.

5.  There are some particles which are always at rest.

Explanation:

In stationary waves $[\text{y}(\text{x, t})=\text{a}\sin\text{kx}\cos\omega\text{t}]$ the particles between two nodes vibrates with different amplitude which increases fron nodes.

The amplitude of a particle will remain constant a cos kx, but varies with $\lambda$

$\because\text{k}=\frac{2\pi}{\lambda}.$ Hence verifies the option (a)

particles between two nodes are in same phase i.e., motion of particles between two nodes will be either upward or downward and crosses the mean position at same time. Hence the verifies option (b)

Hence the reject option (c)

As the particles at nodes are rest so energy does not transfer verifies option (d)

The amplitude of particles at nodes has amplitude zero verifies option (e)

3. Inversly on the square root of density.
4. Directly on the square root of bulk modulus of the medium.

Explanation:

Speed od sound wave in fluid of bulk modules k and density $\rho$ is given by $\text{v}=\sqrt{\frac{\text{k}}{\rho}}$

so $\text{v}=\sqrt{\text{k}}$ (if $\rho$ is constant)

And $\text{v}=\sqrt{\frac{1}{\rho}}$ (if k is constann)

so verifies the option (c) and (d).

1. 1.47Hz

Explanation:

T = 4 × Time from max. displacement to zero displacement.

$=4\times0.170=0.68\text{s}$

$\text{n}=\frac{1}{\text{T}}=\frac{1}{0.68}$

$=1.47\text{Hz}$

2. $2$

Explanation:

When listener alone is moving towards the source.

$\text{n}'=\frac{(\nu+\nu_\text{L})\text{n}}{\nu}$

$\therefore 5.5=\Big(\frac{\nu+\nu_\text{A}}{\nu}\Big)5$

and $6.0=\Big(\frac{\nu+\nu_\text{B}}{\nu}\Big)5$

Solving (i) and (ii), we get

$\frac{\nu_\text{B}}{\nu_\text{A}}=2$

1. The wave is travelling from right to left.
2. The speed of the wave is 20m/ s.
3. Frequency of the wave is 5.7Hz.

Explanation:

The standard from of a wave propagated from left to right i. e., in+ ve direction

$\text{y}=\text{a}\sin(\omega\text{t}-\text{k}\text{x}+\phi)$ and

$\text{y}=3.0\sin\Big(36\text{t}+0.018\text{x}+\frac{\pi}{4}\Big)$ (given)

1. As in given equation x is in positive sigh so given wave travalling from right to left verifies option (a)
2. $\text{k}=0.018=\frac{2\pi}{\lambda}=0.018\Rightarrow\lambda=\frac{2\pi}{0.018}$

$\therefore\text{v}=\text{v}\lambda=\frac{18}{\pi}\times\frac{2\pi}{0.018}=2000\text{cm}/ \ \text{s}=20\text{m/ s}$

Verifies the option (b)

3. $\omega=36$ or $2\pi\text{n}=36$ or $\text{v}=\frac{36}{2\pi}=\frac{18}{3.14}=5.7\text{Hz}$

Verifies option (c) n = 5.7Hz

2. $2\text{a}\sin(\text{kx})$

Explanation:

We have incident wave $\text{y}_1=\text{a}\sin(\text{kx}-\omega\text{t})$

So the reflected wave is $\text{y}_2=\text{a}\sin(\text{kx}+\omega\text{t})$

From principle of superposition, the standing wave equation obtained after superimposing y₁ and y₂,

we get

$\text{y(x, t)}=2\text{a}\sin\text{kx}\cos\omega\text{t}$

Thus, the resultant amplitude is

$\text{A(x)}=2\text{a}\sin\text{kx}$

4. $\frac{\pi}{4}\text{m/s}$

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

1. $8\text{m/s}$

Explanation:

Comparing with the standard wave equation

$\text{y}=\text{r}\sin \Big[\frac{2\pi\text{x}}{\lambda}-\frac{2\pi\text{t}}{\text{T}}-\phi\Big], \text{we get}$

$\frac{2\pi}{\lambda}=0.5\pi,\lambda =4\text{m},\frac{2\pi}{\text{T}}=4\pi,\text{T}=0.5\text{s}$

Speed of wave = 8m/s

4. $\frac{\nu}{3}$

Explanation:

The frequency of sound incident on the cliff.

$\text{f}_1=\frac{\nu}{\nu-\nu'}\text{f}$

where v' is velocity of car (source)

For the sound reflected from the cliff, driver is the listener, moving towards the source

$\therefore$ Frequency of sound heard, $\text{f}_2=\frac{\nu+\nu'}{\nu}\text{f}_1$

$\text{f}_2=\frac{\nu+\nu'}{\nu}\times\frac{\nu}{\nu-\nu'}\text{f}=\frac{\nu+\nu'}{\nu-\nu'}\text{f}$

As $\text{f}_2=2\text{f}$

$\therefore \frac{\nu+\nu'}{\nu-\nu'}=2=\nu'=\frac{\nu}3{}$

1. The frequency of sound as heard by an observer standing on the platform is 400Hz.
2. The speed of sound for the observer standing on the platform is 350m/ s.

Explanation:

As the disatance between listener and source does not change so frequency of sound does not change the as heard listener. i.e.,he heard $\text{v}_0=400\text{Hz}.$ verifies option(a)

$\text{v}_0=400\text{Hz}$ frequency of source of sound.

Velocity of wind $\text{v}_\text{w}=10\text{m/ s}$ from source of listenr.

Speed of sound in still air $= \text{v}_\text{s}=340\text{m/ s}$

As the listenner is standing on platfrom.

Speed od sound with respect to listener $\text{v}_\text{s}+\text{v}_\text{w}=340+10=350\text{m/ s}$

Verifies the option (b)

Reject the option (c)and (d) as it constant $\text{v}_0=400\text{Hz}$

2. $\text{y}=-0.4\sin2\pi\Big(\text{t}+\frac{\text{x}}{2}\Big)$

Explanation:

After reflection of wave changes by phase 180°

$\text{y}_\text{i}=0.6\sin2\pi\Big[\text{t}+\frac{\text{x}}{2}\Big]$

$\text{y}_\text{r}\Big(\frac{2}{3}\times0.6\Big)\sin2\pi\Big[\pi+\text{t}+\frac{\text{x}}{2}\Big]$

$\text{y}_\text{r}=-0.4\sin2\pi\Big(\text{t}+\frac{\text{x}}{2}\Big).$ Hence verifies the option (b).

2. 292c.p.s

Explanation:

Known frequency of A, n₁ = 288c.p.s.

Number of beats/ sec m = 4

$\therefore$ Unknown freq. of B, n₂ = $288\pm4=292 \text{ or }284$

on loading B, number of beats/ sec decreases to 2.

Therefore, m is +.

$\therefore $ n₂ = 288 + 4 = 292c.p.s.

2. It represents a stationary wave of frequency 60Hz.
3. It is the result of superposition of two waves of wavelength 3m, frequency 60Hz each travelling with a speed of 180m/ s in opposite direction.

Explanation:

We know thet standard equation of stationary wave is $\text{y}(\text{x, t})=\text{a}\sin(\text{kx})\cos(\omega\text{t})$ and given equation is $\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\Big)\cos[9120\pi)\text{t})$

2. Comparring both equation $\omega=120\pi$

$2\pi\text{v}=120\pi$ or $\text{v}=\frac{120}{2}=60\text{Hz}$

Verifies the option (b).

3. $\frac{2\pi}{\lambda}=\text{k}$ from $\text{k}=\frac{2\pi}{3}$

$\therefore\frac{2\pi}{\lambda}=\frac{2\pi}{3}$

$\Rightarrow\lambda=3\text{m},\ \text{v}=60\text{Hz}$

speed $\text{v}=\text{v}\lambda=60\times3=180\text{m/ s}.$ Hence, verifies the option (c).

1. $\frac{1}{\text{n}_1-\text{n}_2}$

Explanation:

Time interval between two successive maxima = time interval between two successive beats

$=\frac{1}{\text{n}}=\frac{1}{\text{n}_1-\text{n}_2}$

1. $\nu$

Explanation:

Velocity of sound is independent of frequency. Therefore, it is same $(\nu)$ for frequency n and 4n.

1. 4 × 10^-4m

Explanation:

$\text{n}=4.2\text{MHz}=4.2\times10^6\text{Hz}$

$\nu=1.7\text{km s}^{-1}=1.7\times10^3\text{m s}^{-1}$

$\lambda =\frac{\nu}{\text{n}}=\frac{1.7\times10^3}{4.2\times10^6}$

$=4\times10^{-4}\text{m}$

1. 0.05s

Explanation:

​​​​​​​$\text{n}=\frac{1}{27}\sqrt{\frac{\text{T}}{\text{m}}}=\frac{1}{2\times0.4}\sqrt{\frac{1.6}{\frac{10^{-2}}{0.4}}}=10\text{Hz}$

$\therefore$ Time period t $=\frac{1}{\text{n}}=\frac{1}{10}=0.1\text{s}$

As distance between two fixed ends $=\frac{\lambda}{2}$

$\therefore$ for constructive interference, min value of time

$\Delta \text{t}=\frac{\text{t}}{2}=\frac{0.1}{2}=0.05\text{s}$

3. $2\pi$

Explanation:

When wave passes from one medium to another its frequency (v) does not chang but its velocity and wavelength changs.

$\text{v}=\text{v}\lambda$ or $\text{v}=\frac{\text{v}}{\lambda}$

$\frac{\text{v}}{\lambda}=\frac{2\text{v}}{\lambda_2}\Rightarrow\lambda_2=2\lambda.$

Hence verifies the correct option (c)

3. 0.1 unit

Explanation:

The given equation is $\text{y}=0.0015\sin(62.4\text{x}+316\text{t}),$ Compare it with the standard equation.

$\text{y}=\text{r}\sin\Big(\frac{2\pi}{\lambda}\text{x}+\frac{2\pi\text{t}}{\text{T}}\Big)$

$\frac{2\pi}{\lambda}=62.4$

$\lambda=\frac{2\pi}{62.4}=\frac{2\times3.14}{62.4}=0.1\text{ unit}$

1. Same frequency.

2. Same phase. 

4. Different amplitude.

Explanation:

The frequencies of all particles are same, verifies the option (a).  

particles between any two consecutive nodes vibrates either upside or downside having sameb phase $120\pi\text{t}$ ay atime, verifies the option (b)

particles have ditternt energies. so rejects the option (c)

As the amplitude of different particles are diffrent between two nodes energy (E)  $\propto\text{A}^2.$ verifies the option (d)

1. $\frac{10}{9}$

Explanation:

When source is moving towards a stationary listener, apparent frequency.

$\text{f}'=\frac{\nu\times\text{f}}{\nu-\nu_\text{s}}=\frac{\nu}{\nu-\frac{\nu}{10}}\text{f}=\frac{10}{9}\text{f}$

$\frac{\text{f}'}{\text{f}}=\frac{10}{9}$

2. 2.5Hz, 0.4s

Explanation:

Comparison with the standard equation shows that

$\frac{2\pi}{\text{T}}=5\pi,\text{T}=\frac{2\pi}{5\pi}=0.4\text{s}$

$\text{n}=\frac{1}{\text{T}}=\frac{1}{.4}=2.5\text{Hz}$

1. $400\text{Hz}$

Explanation:

Here, $\text{v}=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\text{m}}}$

When tension is increased by 3T

$\text{v}'=\frac{1}{2\text{l}}\sqrt{\frac{(\text{T}+3\text{T})}{\text{m}}}=2\text{v}$

$\text{v}'=2\times200=400\text{Hz}$

1. Amplitude of all the particles is equal.
2. Particles of the medium executes S.H.M.
3. Wave velocity depends upon the nature of the medium.

Explanation:

During propagation of mechanical wave each particles displaces from zero to maximum i.e., upto amplitude So amplitude of each particle is equal. Verifies the option (b).

For progressive wave medium particles oscillates about their mean position in which restoring force $\text{F}\propto(-\text{y}).$ So motion of medium particles is simple harmonic motion. So verifies the option(c).For progressive wave propagating in a medium of density $(\rho)$ and Bulk modulus k the velocity $(\nu)$

As the depends on k and $(\rho)$ and k, $(\rho)$ are different for different medium so of wave depends on nature of medium, hence, verifies the option (d).

1. $\frac{2\text{u}}{\lambda}$

Explanation:

Number of extra waves received/ sec. $=\pm\frac{\text{u}}{\lambda}$

$\therefore$ Number of beats/ sec. $=\frac{\text{u}}{\lambda}-\Big(\frac{-\text{u}}{\lambda}\Big)=\frac{2\text{u}}{\lambda}$

4. 93ms^-1

Explanation:

Linear mass density = 6.9 × 10^-3 kg m^-1

Tension, T = 60N Thus, speed of wave on the wire is given by

$\nu=\sqrt{\frac{\text{T}}{\mu}}=\sqrt{\frac{60\text{N}}{6.9\times10^{-3}\text{kg m}^{-1}}}$

$=93\text{ms}^{-1}$

3. Wavelength of sound waves.

Explanation:

Speed of sound wave in amedium $\text{v}=\sqrt{\frac{\gamma\ \text{RT}}{\text{M}}}.$ here $\gamma,\text{R}$ and M are constent.

Hence, $\text{v}\propto\sqrt{\text{T}}$ (where T is tempreature of the medium)

It mence when temprature changes, speed also changes.

As, $\text{v}=\text{f}\lambda,$ where f is frequency and $\lambda$ is wavelength.

As frequency (f) remains fixed, $\text{v}\propto\lambda$ or $\lambda$ hence wavelength $(\lambda)$changes.

Hence verifies the option (c).

2. $\text{x}+\frac{2\text{n}\pi}{\text{k}}$

Explanation:

$\text{y}(\text{x}, 0)=\text{a}\sin\text{kx}=\text{a}\sin (\text{kx}+2\text{n}\pi)$

$=\text{a}\sin\text{k}\Big(\text{x}+\frac{2\text{n}\pi}{\text{k}}\Big)$

$\Rightarrow $ The displacement at points x and $\Big(\text{x}+\frac{2\text{n}\pi}{\text{k}}\Big)$

are the same where, n = 1, 2, 3,......

4. $2\text{T}$

3. Increases with increase in humidity.

Explanation:

Speed of sound (longitudinal) wave in air is $\text{v}=\sqrt{\frac{\lambda\text{p}}{\text{p}}}.$ the density of water vapours is less (rises up) than the air so on increasing humidity, the density of medium decrease in turn increases the speed of sound in air by $\text{v}=\propto\frac{1}{\sqrt{\text{p}}}$ (relation).

Hence verifies the option (c).

2. $15\text{ms}^{-1}$

Explanation:

Here, f = 9500Hz, $\nu_\text{s}=?$, v = 300ms^-1, f' = 1000Hz

As the source is approaching a stationary listener,

$\therefore \text{f}'=\frac{\nu\times\text{f}}{\nu-\nu_\text{s}}$

$1000=\frac{300\times9500}{300-\nu_\text{s}}$

$285=300=\nu_\text{s}$

$\nu_\text{s}=300-285=15\text{ms}^{-1}$