Questions Related to physics

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

Consider the following two equations $L=I\omega$ and $ \dfrac { dL }{ dt } =\Gamma $. In noninertial frames :

  1. both A and B are true

  2. A is true but B is false

  3. B is true but A is false

  4. both A and B are false.

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In non-inertial frames, the relation L = I*omega is generally not valid due to the changing nature of the moment of inertia or frame-dependent definitions of angular momentum. However, the torque equation dL/dt = Gamma is a fundamental law of motion that holds in inertial frames, but in non-inertial frames, pseudo-torques must be included.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation $y = a \sin^2 \left(2 \pi nt - \dfrac{2\pi x}{\lambda}\right)$ represents a wave with

  1. Amplitude $a$, frequency $n$ and wavelength $\lambda$
  2. Amplitude $a$, frequency $2n$ and wavelength $2\lambda$
  3. Amplitude $a/2$, frequency $2n$ and wavelength $\lambda$
  4. Amplitude $a/2$, frequency $2n$ and wavelength $\lambda/2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using sin^2(theta) = (1 - cos(2*theta))/2, the equation becomes y = a/2 - (a/2)cos(4*pi*n*t - 4*pi*x/lambda). This represents a wave with amplitude a/2, frequency 2n, and wavelength lambda/2 (since k = 4*pi/lambda = 2*pi/lambda_new).

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The speed of the wave travelling on the uniform circular hoop of string, rotating clockwise in absence of gravity with tangential speed $v _0$, is :

  1. $v=v _0$
  2. $v=2v _0$
  3. $v=\dfrac{v _0}{\sqrt 3}$
  4. $v=\dfrac{v _0}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a string rotating in a circle, the wave speed relative to the string is v0. In the absence of gravity, the speed of a transverse wave on a rotating hoop is equal to the tangential speed v0.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation $y =A\cos^2\left(2\pi\, nt -2\pi \dfrac{x}{\lambda}\right)$ represents a wave with

  1. amplitude $A/2$, frequency $2n$& wavelength $\lambda/2$
  2. amplitude $A/2$, frequency $2n$& wavelength $\lambda$
  3. amplitude $A$, frequency $2n$& wavelength $2\lambda$
  4. amplitude $A$, frequency $n$& wavelength $\lambda$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using cos^2(theta) = (1 + cos(2*theta))/2, the equation becomes y = A/2 + (A/2)cos(4*pi*n*t - 4*pi*x/lambda). This corresponds to amplitude A/2, frequency 2n, and wavelength lambda/2.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation of a wave is given by 
$ Y\quad =\quad A\quad sin\quad \omega \left( \frac { x }{ v } -k \right)  $
Where $ \omega $ is the angular velocity and v is the linear velocity.The dimensions of K is

  1. LT

  2. T

  3. $ T^{-1} $
  4. $ T^2 $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the argument omega(x/v - k), the dimensions of omega are T^-1. For the argument to be dimensionless, (x/v - k) must have dimensions of T. Since x/v is distance/velocity = time, k must also have dimensions of time.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation of a progressive wave is $Y= a sin(200 t-x)$, where x is in meter and t is in second. The velocity of wave is

  1. $200 $ m/sec
  2. $100 $ m/sec
  3. $50 $ m/sec
  4. None

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The wave equation is y = a sin(200t - x). The standard form is y = a sin(omega*t - k*x). Here omega = 200 and k = 1. Wave velocity v = omega/k = 200/1 = 200 m/s.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

The equation of a wave travelling on a stretched string is :
$y=4\sin 2\pi \left(\dfrac{t}{0.02}-\dfrac{x}{100}\right)$
Here $x$ and $y$ are in $cm$ and $t$ is in second. the relative deformation amplitude of medium is :

  1. $0.02\pi$
  2. $0.08\pi$
  3. $0.06\pi$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The relative deformation (strain) is dy/dx. y = 4 sin(2pi*t/0.02 - 2pi*x/100). dy/dx = 4 * (-2pi/100) * cos(...) = -0.08pi * cos(...). The amplitude of this is 0.08pi.

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

A source oscillates with a frequency 25 Hz and the wave propagates with 300 m/s. Two points A and B are located at distances 10 m and 16 m away from the source. The phase difference between A and B is 

  1. $\displaystyle \frac{\pi}{4}$
  2. $\displaystyle \frac{\pi}{2}$
  3. $\pi$
  4. $2 \pi$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Wavelength of the wave=$\lambda=\dfrac{v}{\nu}=\dfrac{300}{25}=12m$

Distance between the two points=$16m-10m=6m=\dfrac{\lambda}{2}$
$=\dfrac{2\pi}{\lambda}\dfrac{\lambda}{2}=\pi$

Multiple choice problems on properties of waves terms and defination used in wave motion oscillation and waves waves physics

Two simple harmonic motions are represented by the equations 
$y _1=10\sin \left(3\pi t+\dfrac{\pi}{4}\right)$
and $y _2=5(3\sin 3\pi t+\sqrt 3 \cos 3\pi t)$ Their amplitudes are in the ratio of :

  1. $\sqrt 3$
  2. $1/\sqrt 3$
  3. $2$
  4. $1/6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

y1 = 10 sin(3pi*t + pi/4), amplitude = 10. y2 = 5(3 sin(3pi*t) + sqrt(3) cos(3pi*t)). Using R = sqrt(A^2 + B^2 + 2AB cos(phi)), y2 = 5 * sqrt(3^2 + sqrt(3)^2) * sin(...) = 5 * sqrt(9+3) = 5 * sqrt(12) = 10 * sqrt(3). Ratio = 10 / (10 * sqrt(3)) = 1/sqrt(3).