Tag: wave motion

Questions Related to wave motion

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

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

From the given standing wave equation y = 5 sin(pi x / 3) cos(40 pi t), the wave number k = pi / 3. The separation between two consecutive nodes is equal to half of the wavelength, lambda / 2. Since lambda = 2 pi / k = 2 pi / (pi / 3) = 6 cm, the separation between consecutive nodes is 6 / 2 = 3 cm.

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

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

Given y = 4 sin(πx/15) cos(96π/t) - note this is TYPO: should be cos(96πt). Maximum displacement occurs when cos(96πt) = ±1. At x=5 cm, amplitude = 4 sin(π×5/15) = 4 sin(π/3) = 4 × √3/2 = 2√3 cm.

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.