Tag: physics

Questions Related to physics

Small amplitude progressive waves in a stretched string have a speed of 100 cm/s and frequency 100 Hz. The phase difference between two points 2.75 cm apart on the string, in radians is 

  1. $\dfrac { \pi }{ 4 } $

  2. $\dfrac { 3\pi }{ 4 } $

  3. $0$

  4. $\dfrac { 11\pi }{ 4 } $


Correct Option: D

A tension in wire is 40N and 10 m of wire has a mass of 0.01 kg . The speed of transverse waves in m/s in the wire is :

  1. 200

  2. 80

  3. 300

  4. 180


Correct Option: A
Explanation:

We know, Speed of transverse wave $(v) = \sqrt{\dfrac{T}{\mu}}$


where, T = Tension = 40N and  $\mu = $ mass per unit length = $\dfrac{0.01}{10} = 10^{-3}\; kg/m$ 

$\Rightarrow v = \sqrt{\dfrac{40}{10^{-3}}} = 200 m/s$

Therefore, A is correct option.

A string of mass $2.5\ kg$ is under a tension of $200\ N$. The length of the stretched string is $20.0\ m$. If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in

  1. One second

  2. $0.5$ second

  3. $2\ seconds$

  4. Data given is insufficient


Correct Option: B

The  equation of a transverse wave travel on a rope is given   y = 10 sin $\pi$(0.01x - 2.00t) where y and x in cm and t in seconds.The maximum transverse  speed  of a particle in the rope about 

  1. 62.8 cm / s

  2. 75 cm / s

  3. 100 cm / s

  4. 121 cm / s


Correct Option: A

A wave represented by equation $y = 2(mm) \, sin \, [4 \pi (sec^{-1}) t - 2 \pi (m^{-1}) X]$ is superimposed with another wave $y = 2 (mm) sin [4 \pi (sec^{-1}) t + 2 \pi (m^{-1}) x + \pi/3]$ on a tight string.
Phase difference between two particles with are located at $x _1 = 1/7$ and $x _2 = 5/12$ is :

  1. $0$

  2. $\dfrac{5 \pi}{6}$

  3. $\pi$

  4. $\dfrac{5 \pi}{3}$


Correct Option: C
Explanation:

${y _1} = 2\sin \left[ {4\pi t - 2\pi x} \right]$

${y _2} = 2\sin \left[ {4\pi t - 2\pi x} \right]$
$y = {y _1} + {y _2} = 2\left[ {2\sin 4\pi t\,\,\cos 4\pi x} \right]$
$ = 4\sin 4\pi t\cos 4\pi x$
$y = 4\cos 4\pi \sin 4\pi t$
Amp pass$\left| {4\cos 4\pi x} \right| = 4$
$ \Rightarrow \cos 4\pi x =  \pm 1$
$ = 0,\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4},\frac{5}{4},\frac{6}{4}.......$
$\therefore for\,{x _1} = \frac{1}{9}and\,{x _2} = \frac{5}{{12}}$
hence,
phase difference is$\pi$

A travelling wave on a string is given by $y = A$ $A \sin \left[ \alpha x + \beta t + \cfrac { \pi } { 6 } \right]$ The displacement and velocity of oscillation of a point $\alpha =$ $0.56 / \mathrm { cm } , \beta = 12 / \mathrm { sec }$ $A = 7.5 \mathrm { cm } , x = 1$ $\mathrm { cm }$ and $\mathrm { t } = 1 \mathrm { s }$ is

  1. $4.6 \mathrm { cm } , 46.5 \mathrm { cm } s ^ { - 1 }$

  2. $3.75 \mathrm { cm } , 77.94 \mathrm { cm } \mathrm { s } ^ { - 1 }$

  3. $1.76 \mathrm { cm } , 7.5 \mathrm { cms } ^ { - 1 }$

  4. $7.5 \mathrm { cm } , 75 \mathrm { cm } \mathrm { s } ^ { - 1 }$


Correct Option: B
Explanation:

$\begin{array}{l} y=7.5\sin  \left[ { 0.56\alpha +12t+\dfrac { \pi  }{ 6 }  } \right]  \ at\, \, x=1 \ & \, \, t=1 \ y=7.5\sin  \left[ { 0.56+12+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 12.56+\dfrac { \pi  }{ 6 }  } \right]  \ =7.5\sin  \left[ { 4\pi +\dfrac { \pi  }{ 6 }  } \right]  \ =3.75\, \, cm \end{array}$

$\therefore$ Option $B$ is correct .

A sine wave is travelling in a medium. The minimum distance between the two particles. always having same speed is 

  1. $\lambda / 4$

  2. $\lambda / 3$

  3. $\lambda / 2$

  4. $\lambda $


Correct Option: C
Explanation:

In a sine wave particles that are separated by a distance of odd multiple of half

the wave length move with same speed and but in opposite direction. 
The minimum separation is $\frac{\lambda }{2}$
Option C.

A kite flying at a height h meter has r meter of string paid out at a time of t sec . If the kite moves horizontally with constant velocity v meter/sec then the at which the string is paid out is

  1. $\sqrt{\left ( r^2 h^2 \right )v}$ mt/sec

  2. $\dfrac{v\sqrt{\left ( r^2 h^2 \right )}}{r}$ mt/sec

  3. $\dfrac{r\sqrt{\left ( r^2 h^2 \right )}}{v}$ mt/sec

  4. $\dfrac{\sqrt{\left ( r^2 h^2 \right )}}{rv}$ mt/sec


Correct Option: C

The equation of standing wave in a stretched string us given by $y=5\sin\left(\cfrac{\pi x}{3}\right)\cos(40\pi t)$, where $x$ and $y$ are in cm and $t$ in seconds. The seperation between two consecutive nodes is (in cm)

  1. $1.5$

  2. $3$

  3. $6$

  4. $4$


Correct Option: A

The vibration of string of length 60 cm fixed at both ends are represented by the equations $ y=4 sin ( \pi x / 15 ) cos ( 96 \pi / t ) $ where x and y are in cm and t in s. the maximum displacement at x=5 cm is 

  1. $ 2 \sqrt 3 cm $

  2. $4 cm$

  3. $zero$

  4. $ 4 \sqrt 2 cm $


Correct Option: A