The amplitude of a wave represented by displacement equation
$y = \frac{1}{{\sqrt a }}\,\sin \,\omega t \pm \frac{1}{{\sqrt b }}\,\cos \,\omega t$ will be
Medium
Download our app for free and get started
Ideal equation
$y=\sqrt{\left(\frac{1}{\sqrt{a}}\right)^{2}+\left(\frac{1}{\sqrt{b}}\right) \sin (\omega t+\phi)}$
$\mathrm{y}=\left(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}\right) \sin (\omega \mathrm{t}+\phi)$
amplitude $A=\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a+b}{a b}}$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is $x(t)$ vs. $p(t)$ curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative. $Image$View Solution
$1.$ The phase space diagram for a ball thrown vertically up from ground is
mcq $Image$
$2.$ The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and $E_1$ and $E_2$ are the total mechanical energies respectively. Then $Image$
$(A)$ $ E_1=\sqrt{2} E_2$ $(B)$ $ E_1=2 E_2$
$(C)$ $ E_1=4 E_2$ $(D)$ $ E_1=16 E_2$
$3.$ Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is $Image$
mcq $Image$
Give the answer question $1,2$ and $3.$
- 2The motion of a particle represented by $y\ =$ $\sin \omega t - \cos \omega t$ isView Solution
- 3A $15 \,g$ ball is shot from a spring gun whose spring has a force constant of $600 \,N/m$. The spring is compressed by $5 \,cm$. The greatest possible horizontal range of the ball for this compression is .... $m$ ($g = 10 \,m/s^2$)View Solution
- 4View SolutionTo make the frequency double of an oscillator, we have to
- 5Displacement-time equation of a particle executing $SHM$ is $x\, = \,A\,\sin \,\left( {\omega t\, + \,\frac{\pi }{6}} \right)$ Time taken by the particle to go directly from $x\, = \, - \frac{A}{2}$ to $x\, = \, + \frac{A}{2}$ isView Solution
- 6Two springs, of force constants $k_1$ and $k_2$ are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f$ If both $k_1$ and $k_2$ are made four times their original values, the frequency of oscillation becomesView Solution
- 7A block of mass $200\, g$ executing $SHM$ under the influence of a spring of spring constant $K=90\, N\,m^{-1}$ and a damping constant $b=40\, g\,s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is ...... $s$ (Given $ln\,\frac{1}{2} = -0.693$)View Solution
- 8Two pendulums of length $121\,cm$ and $100\,cm$ start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :View Solution
- 9The scale of a spring balance reading from $0$ to $10 \,kg$ is $0.25\, m$ long. A body suspended from the balance oscillates vertically with a period of $\pi /10$ second. The mass suspended is ..... $kg$ (neglect the mass of the spring)View Solution
- 10A mass m oscillates with simple harmonic motion with frequency $f = \frac{\omega }{{2\pi }}$ and amplitude A on a spring with constant $K$ , thereforeView Solution