In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is $0.170\,$second. The frequency of the wave is .... $Hz$
AIIMS 2001,AIPMT 1998, Medium
Download our app for free and get started
(a) Time required for a point to move from maximum displacement to zero displacement is
$t = \frac{T}{4} = \frac{1}{{4n}}$
==> $n = \frac{1}{{4t}} = \frac{1}{{4 \times 0.170}} = 1.47\,Hz$
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
- 1A disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should beView Solution
- 2$A$ particle of mass m is constrained to move on $x$ -axis. $A$ force $F$ acts on the particle. $F$ always points toward the position labeled $E$. For example, when the particle is to the left of $E, F$ points to the right. The magnitude of $F$ is a constant $F$ except at point $E$ where it is zero. The system is horizontal. $F$ is the net force acting on the particle. The particle is displaced a distance $A$ towards left from the equilibrium position $E$ and released from rest at $t = 0.$ Find minimum time it will take to reach from $x = - \frac{A}{2}$ to $0$.View Solution
- 3View SolutionThe phase difference between displacement and acceleration of a particle in a simple harmonic motlon is
- 4The displacement of a particle executing SHM is given by $x=10 \sin \left(\omega t+\frac{\pi}{3}\right) \mathrm{m}$. The time period of motion is $3.14 \mathrm{~s}$. The velocity of the particle at $\mathrm{t}=0$is_________. $\mathrm{m} / \mathrm{s}$.View Solution
- 5The time period of oscillations of a simple pendulum is $1$ minute. If its length is increased by $44 \%$. then its new time period of oscillation will be ......... $s$View Solution
- 6As per given figures, two springs of spring constants $K$ and $2\,K$ are connected to mass $m$. If the period of oscillation in figure $(a)$ is $3 s$, then the period of oscillation in figure $(b)$ will be $\sqrt{ x }$ s. The value of $x$ is$.........$View Solution
- 7A particle executes linear simple harmonic motion with an amplitude of $2\, cm$. When the particle is at $1\, cm$ from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds isView Solution
- 8A particle executes simple harmonic motion represented by displacement function as $x(t)=A \sin (\omega t+\phi)$View Solution
If the position and velocity of the particle at $t=0\, {s}$ are $2\, {cm}$ and $2\, \omega \,{cm} \,{s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \,{cm}$ where the value of $x$ is ..... .
- 9A simple harmonic motion is represented by $y\, = 5\,(\sin \,3\pi t\, + \,\sqrt 3 \,\cos \,3\pi t)\,cm$ The amplitude and time period of the motion areView 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