Tag: wave motion

Quizzes (25) Questions (122)

Questions Related to wave motion

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

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45 Hz$. The mass of the wire is $3.5 \times 10^{-2}kg$ and its linear mass density is $4.0 \times 10^{-2} kgm^{-1}$. What is the speed of a transverse wave on the wire?

  1. $69 \ ms^{-1}$
  2. $79 \ ms^{-1}$
  3. $89 \ ms^{-1}$
  4. $99 \ ms^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a wire vibrating in its fundamental mode, the frequency f = v / (2L). However, we can use the relation v = sqrt(T/mu). Given the mass M = 0.035 kg and linear density mu = 0.04 kg/m, the length L = M/mu = 0.875 m. The fundamental frequency f = v / (2L) = 45 Hz, so v = 45 * 2 * 0.875 = 78.75 m/s, which rounds to 79 m/s.

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

A person observe two points on a string as a travelling wave passes them. The points are at $x _ { 1 } = 0$ and $x _2 = 1m$. The transverse motions of the two points are found to be as follows: $y _ { 1 } = 0.2 \sin 3 \pi t$
$y _ { 2 } = 0.2 \sin ( 3 \pi t + \pi/8 )$ What is the frequency in Hertz?

  1. $1.5 Hz$
  2. $3 Hz$
  3. $4.5 Hz$
  4. $1 Hz$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The transverse motion is given by y = A sin(omega * t + phi). Comparing y1 = 0.2 sin(3 * pi * t) with the standard form, omega = 3 * pi. Since omega = 2 * pi * f, we have 3 * pi = 2 * pi * f, which gives f = 1.5 Hz.

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

If $n,2n,3n$ are the fundamental frequencies of the three segments into which a string is divided by placing required number of bridges below it. If $n _0$ is the fundamental frequency of the string, then 

  1. $n _0=3n$
  2. $n _0=6n$
  3. $n _0=\dfrac{3n}{5}$
  4. $n _0=\dfrac{6n}{11}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The fundamental frequency of a string is f = v / (2L). When divided into segments of lengths L1, L2, L3, the frequencies are f1 = v / (2L1) = n, f2 = v / (2L2) = 2n, f3 = v / (2L3) = 3n. The total length L = L1 + L2 + L3 = v/(2n) + v/(4n) + v/(6n) = (6+3+2)v / 12n = 11v / 12n. The fundamental frequency of the whole string is f0 = v / (2L) = v / (2 * 11v / 12n) = 6n / 11.

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

A spring of force constant K is first stretched by distance a from its natural length and then future by distance b. The work done in stretching the part b is

  1. $\dfrac{1}{2}$Ka(a-b)
  2. $\dfrac{1}{2}$Ka(a+b)
  3. $\dfrac{1}{2}$Kb(a-b)
  4. $\dfrac{1}{2}$Kb(2a+b)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Work done by spring in its natural length$=\cfrac{1}{2} \times k \times x^{2}= \cfrac{1}{2} \times k \times a^{2}$

So, total work$=\cfrac{1}{2}k(a+b)^{2}$
for work done for stretching 'b'
$\cfrac { 1 }{ 2 } \times k\times (a+b)^{ 2 }-\cfrac { 1 }{ 2 } \times k\times a^{ 2 }=\cfrac { 1 }{ 2 } \times k\times b\times (2a+b)$

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

Two tuning forks when sounded together produce 5 beat per second. The first tuning fork is in resonance with 16.0 cm wire of a sonometer and the second is in resonace with 16.2 cm wire of the same sonometer. The frequencies of the tuning forks are

  1. 100 Hz,105 Hz

  2. 20 Hz,205 Hz

  3. 300 Hz,305 Hz

  4. 400 Hz,405 Hz
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The frequency of a sonometer wire is inversely proportional to its length, f proportional to 1/L. Thus, f1 * L1 = f2 * L2, meaning f1 * 16.0 = f2 * 16.2. Also, the beat frequency is f2 - f1 = 5. Solving these simultaneous equations: 16.0 f1 = (f1 + 5) * 16.2, giving 16.0 f1 = 16.2 f1 + 81, so 0.2 f1 = 81, leading to f1 = 400 Hz and f2 = 405 Hz.

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

The equation of wave in string is $\displaystyle y = 20\sin \frac{\pi x}{2} \cos 40\pi t$ in metre. The speed of the wave is 

  1. $Zero$
  2. $80\, m/s$
  3. $320\, m/s$
  4. $160\, m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\large \begin{array}{l} Here, \ y=20\sin  \frac { { \pi x } }{ 2 } \cos  40\pi t-----(i) \ compare\, with\, eqution,\, (i)\,  \ \Rightarrow y=2r\, \, \sin  \frac { { 2\pi  } }{ \lambda  } \, \times \, \, \cos  \frac { { 2\pi  } }{ \lambda  } Vt \ Now, \ \Rightarrow \frac { { 2\pi  } }{ \lambda  } =\frac { \pi  }{ 2 } \, \, and\, \, \frac { { 2\pi  } }{ \lambda  } V=40\pi  \ so, \ \Rightarrow \frac { \pi  }{ 2 } \, \times V=40\pi  \ \therefore \, \, V=80\, m/s \end{array}$

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

A body of mass m is tied to one of a spring and whirled round in a horizontal plane with a constant angular velocity and elongation in the spring is 1 cm. If the angular velocity is doubled, the elongation in the spring becomes 5 cm. The original length of spring is 

  1. 20 cm

  2. 25 cm

  3. 10 cm

  4. 15 cm
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The centripetal force is provided by the spring force: m * omega^2 * (r + x) = k * x, where r is the original length and x is the elongation. For omega, x1 = 1 cm. For 2 * omega, x2 = 5 cm. Thus, m * omega^2 * (r + 1) = k * 1 and m * (2 * omega)^2 * (r + 5) = k * 5. Dividing the equations: 4 * (r + 5) / (r + 1) = 5. Solving for r: 4r + 20 = 5r + 5, so r = 15 cm.

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

A 100 Hz sinusoidal wave is travelling in the positive x-direction along a string with a linear mass density of $3.5\, \times\, 10^{-3}\, kg/m$ and a tension of 35 N. At time t = 0, the point x = 0, has maximum displacement in the positive y direction. Next when this point has zero displacement the slope of the string is $\pi /20$. which of the following expression represent (s) the displacement of string as a function of x (in metre) and t (in second).

  1. $y\, =\, 0.025\, cos\, (200 \pi t\, -\, 2 \pi x)$
  2. $y\, =\, 0.5\, cos\, (200 \pi t\, -\, 2 \pi x)$
  3. $y\, =\, 0.025\, cos\, (100 \pi t\, -\, 10 \pi x)$
  4. $y\, =\, 0.5\, cos\, (100 \pi t\, -\, 10 \pi x)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the wave have the form $y=Asin(\omega t-kx+\phi)$
Since the frequency is $100Hz$, $\omega=2\pi\nu=200\pi$
Speed of the wave=$\sqrt{\dfrac{T}{\mu}}=\dfrac{\omega}{k}$
$\implies k=2\pi$
Since displacement is maximum at (x,t)=(0,0), $sin(0+0+\phi)=1$
$\implies \phi=\dfrac{\pi}{2}$
Thus the wave is $y=Acos(\omega t-kx)$
$Slope=\left|\dfrac{dy}{dx}\right|=Aksin(\omega t-kx)=\dfrac{\pi}{20}$ at $(x,t)=(0,0)$
Thus $Ak=\dfrac{\pi}{20}$
$\implies A=0.025m$
Thus the correct answer is option A.
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

A harmonic oscillator vibrates with amplitude of 4 cm and performs 150 oscillations in one minute. If the initial phase is $45\circ$ and it starts moving away from the equation of motion is 

  1. $\displaystyle 0.04\, sin\, \left ( 5 \pi t\, +\, \frac{\pi}{4} \right )$
  2. $\displaystyle 0.04\, sin\, \left ( 5 \pi t\, -\, \frac{\pi}{4} \right )$
  3. $\displaystyle 0.04\, sin\, \left ( 4 \pi t\, +\, \frac{\pi}{4} \right )$
  4. $\displaystyle 0.04\, sin\, \left ( 4 \pi t\, -\, \frac{\pi}{4} \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of motion of harmonic oscillator is $y=Asin(\omega t+\phi _0)$

It is given that $A=0.04m$
and $\phi _0=54^{\circ}=\dfrac{45}{180}\pi=\dfrac{\pi}{4}$
$\omega=2\pi\nu=2\pi\times \dfrac{150}{60}=5\pi$
Thus $y=0.04sin(5\pi t+\dfrac{\pi}{4})$
Hence the answer is option A.