Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A uniform string of length 20 m & mass 1 Kg is hung vertically.Find the speed of wave at the mid point of the string :-

  1. 20 m/s

  2. 30 m/s

  3. $ 10 \sqrt {2} m/s $

  4. 10 m/s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Tension at midpoint: The weight of the lower half of the string is (m/2) * g = 0.5 * 10 = 5 N. Linear mass density mu = 1 kg / 20 m = 0.05 kg/m. Wave speed v = sqrt(T / mu) = sqrt(5 / 0.05) = sqrt(100) = 10 m/s.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A standing wave of time period T is set up in string clamped between two rigid supports at t=0 antitode is at its maximum displacement A

  1. The energy of a node is equal to energy of an anitode for the first time at t=T/8

  2. The energy of node and antitode becomes equal after every T/2 second.

  3. the displacement of the particle of antinode at $ t= \frac {T}{8} is \sqrt 2 A $

  4. The displacement of the particle of node is zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A man generates a ssmmetrical pulse in a string by moving his hand up and down. At $t = 0$ the how hand mowes downuard: The pulse travels with speed of 3$\mathrm { m } / \mathrm { s }$ on the string $&$ his hands passe 6 in each secand from the mean position. Then the point on the string at a distance 3$\mathrm { m }$ will reach its topper arreme first time at time t=

  1. 0.25 sec.

  2. 1 sec

  3. $\frac { 13 } { 12 } \mathrm { sec }$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Hand completes 6 oscillations per second, so period T = 1/6 s. At t=0, hand moves downward. Pulse travels at 3 m/s. Point at 3 m reaches upper extreme when pulse arrives and phase is maximum. Time = distance/speed + T/4 = 3/3 + (1/6)/4 = 1 + 1/24 = 25/24 ≈ 0.25 s.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A rope of length $L$ and mass $m$ hangs freely from the celling. The velocity of transverse wave as a furcion position $x$ along the rope is proportional to

  1. $x ^ { 0 }$

  2. $\sqrt { x }$

  3. $\frac { 1 } { \sqrt { x } }$

  4. $x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Tension at distance x from the free end is T = mu * g * x. Wave speed v = sqrt(T / mu) = sqrt(g * x). Thus, v is proportional to sqrt(x).

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

If a string is stretched by $\dfrac{L}{20}$ then velocity of wave is $V$. When string is stretched by $\dfrac{L}{10}$ then velocity becomes

  1. $\dfrac{V}{\sqrt 2}$

  2. $V$

  3. $2V$

  4. $\sqrt 2 V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For a stretched string, T = Y * A * (delta_L / L). Wave speed v = sqrt(T / mu) = sqrt(Y * A * delta_L / (L * mu)). Since v is proportional to sqrt(delta_L), doubling the extension delta_L (from L/20 to L/10) increases the velocity by a factor of sqrt(2).

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

Sinusoidal waves 5.00 cm in amplitude are to be transmitted along a string having a linear mass density equal to 4.00 * $10^-2 kg/m$. If the source can deliver a average power of 90 W and the string is under a tension of 100 N,then the highest frequency at which the source can operate is (take $\pi^2 = 10)$:

  1. 45 Hz

  2. 50 Hz

  3. 30 Hz

  4. 62 Hz
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Average power P = 2 * pi^2 * f^2 * A^2 * mu * v. Tension T = 100 N, mu = 0.04 kg/m, so v = sqrt(100 / 0.04) = 50 m/s. P = 90 W, A = 0.05 m. 90 = 2 * 10 * f^2 * (0.05)^2 * 0.04 * 50. 90 = 20 * f^2 * 0.0025 * 2 = 0.1 * f^2. f^2 = 900, so f = 30 Hz.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

The equation of a stationary wave in a string is y =(4) sin[($3.14m^-1$)x] cos ${\omega}t$. (mm)
Select the correct alternative(s).

  1. the amplitude of component waves is 2 mm

  2. the amplitude of component waves is 4 mm

  3. the smallest possible length of string is 0.5 m

  4. the smallest possible length of string is 1.0 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of stationary wave in a string is the amplitude of component waves is $4$mm.

Hence, option $B$ is correct ansnwer.

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A wave represented by a given equation $y (x,t) = a \sin (\omega t - kx)$superimposes on another wave giving a stationary wave having antinode at $x = 0 $ then the equation of the another wave is 

  1. $y = - a \sin (\omega t - kx)$

  2. $y = a \sin (\omega t + kx)$

  3. $y = - a \sin (\omega t + kx)$

  4. $y = - a \cos (\omega t + kx)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} a\sin  \left( { wt-kx } \right) +{ y _{ 1 } }=2a\sin  \cot  \cos  kx \ \Rightarrow y=a\sin  \left( { wt+kx } \right)  \ Ans.\, \, (B) \end{array}$

Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A travelling wave passes  point of observation. At this point, the time interval between successive crests is $0.2\, s$ and 

  1. The wavelength is $5\, m$

  2. The frequency is $5\, Hz$

  3. The velocity of propagation is $5\, m/s$

  4. The wavelength is $0.2\, m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} Accorrding\, \, to\, \, question....................... \ Here, \ Difference\, between\, two\, successive\, crest\, is\, 2p. \ passes\, difference\, (\Delta \varphi )=\frac { { 2\pi  } }{ T }  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, T=time\, { { int } }erval(\Delta t) \ \therefore \, \, \, n=2\pi =\frac { { 2\pi  } }{ T } \times 0.2 \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Rightarrow \frac { 1 }{ T } =5{ \sec ^{ -1 }  } \ \Rightarrow n=5Hz \ So\, the\, correct\, option\, is\, B. \end{array}$