Sinusoidal Wave
Trending Questions
y=Asin(5t−0.05x) in which A , x are in metres and t is in seconds. The value of A is
y=10sin(πt−0.5x), where y and x are in m and time(t) in sec. Find the acceleration of a particle at x=π m at time t=1 sec.
- −10π m/s2
- 5π m/s2
- −10π2 m/s2
- −5π m/s2
- λ2
- λ
- λ4π
- λ2π
The equation of a transverse wave travelling on a rope is given by y=10sin π(0.001x−2.00t) where y and x are in centimeters and t in seconds. The maximum transverse speed of a particle in the rope is about.
121 cm/s
63 cm/s
100 cm/s
75 cm/s
The equation of a transverse wave travelling on a rope is given by y=10sin π(0.001x−2.00t) where y and x are in centimeters and t in seconds. The maximum transverse speed of a particle in the rope is about.
63 cm/s
75 cm/s
100 cm/s
121 cm/s
Why a transverse wave can be either mechanical or non-mechanical wave ?
- y=−0.4sin2π(t+x2)
- y=0.6sin2π(t+x2)
- y=−0.4sin2π(t−x2)
- y=0.4sin2π(t+x2)
Mathematically, it is given by →S=1μ0(→E×→B).
Show the nature of S vs t graph.
- y=x2−t2
- y=log(x2−t2)−log(x−t)
- y=e2xsint
- y=cosx2sint
- λ'<λ
- λ'=λ
- λ'>λ
- None
When a wave travels in a medium, the particle displacement is given by
y(x, t)=0.03 sin π(2t−0.01x)
where y and x are expressed in metres and t in seconds. At a given instant of time, the phase difference between two particles 25 m apart is
π8
π4
π2
π
- 0.5 ms
- 2.4 ms
- 1.9 ms
- 3.9 ms
Figure (15-E2) shows a plot of the transverse displacements of the particles of a string at t = 0 through which a travelling wave is passing in the positive x-direction. The wave speed is 20 cm s−1. Find (a) the amplitude, (b) the wavelength, (c) the wave number and (d) the frequency of the wave.
As the wavelength of a wave in a uniform medium increases, then what will happen with its speed?
Simplify the expression
- nv(R−2)π
- nvπR
- nvR
- nvR(π−2)
y=Asin(kx−ωt+ϕ0)
The term phase is defined as:
- ϕ0
- ϕ0−ωt
- kx+ϕ0
- kx−ωt+ϕ0
- True
- False
Column IColumn II(A) Frequency of wave(Hz)(p) a(B) wavelength of wave(m)(q) b(C) Phase difference between two points 14a m distance apart(r) π(D) Phase difference of a particle after a time interval of 18b s(s) π2(t) None of above
- ABCDqtst
- ABCDtqrs
- ABCDpqrs
- ABCDtspr
- Aω
- Aω2
- Aω3
- Aω4
Find the acceleration of the particle at x=0.85 m and t=0.25 s
[Assume, π2=10 ; ↑ denotes positive y−direction and ↓ denotes negative y−direction ]
- 40 m/s2 ↑
- 50 m/s2 ↓
- 40 m/s2 ↓
- 50 m/s2 ↑
The equation for a standing wave generated by a string, with both ends fixed, is. Figure out the smallest length of the string.
- √2A0
- √3A0
- A0
- 2A0
Figure (16-E12) shows a source of sound moving along the X-axis at a speed of 22 m s−1 continuously emitting a sound of frequency 2.0 kHz which travels in air at a speed of 330 m s−1. A listener Q stands on the Y-axis at a distance of 330 m from the origin. At t = 0, the source crosses the origin P. (a) When does the sound emitted from the source at P reach the listener Q ? (b) What will be the frequency heard by the listener at this instant? (c) Where will the source be at this instant ?
- y(x, t)=(2 cm)sin[(1.68 rad/s)t−(1 cm−1)x]
- y(x, t)=sin[(3.2 rad/s)t−(0.42 cm−1)x]
- y(x, t)=(2 cm)cos[(3.2 rad/s)t−(0.42 cm−1)x]
- y(x, t)=(4 cm)cos[(1.68 rad/s)t−(0.42 cm−1)x]
y1=(2.00 cm) sin (20x−32t) and y2=(2.00 cm) sin (25x−40t) where y1, y2 and x are in centimeters and t is in seconds. What is the phase difference between these two waves at the point x=5.00 cm at t=2 s?
- 516∘
- 258∘
- 333∘
- 412∘