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Question

A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec. The time required by it to move from $x = A$ to $x = \frac{A}{\sqrt{2}}$ is _________ sec.

The correct answer is
5/8

Understanding Simple Harmonic Motion

The position $x$ of a particle in Simple Harmonic Motion (SHM) can be described by the equation: $x(t) = A \cos(\omega t + \phi)$ where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is time, and $\phi$ is the phase constant.

Calculating Angular Frequency

The angular frequency $\omega$ is related to the time period $T$ by: $\omega = \frac{2\pi}{T}$ Given the time period $T = 5$ sec, the angular frequency is: $\omega = \frac{2\pi}{5}$ rad/sec.

Determining Phase and Time

We assume the motion starts from the extreme position $x = A$ at $t=0$. This implies the phase constant $\phi = 0$. The equation of motion simplifies to: $x(t) = A \cos(\omega t)$

We need to find the time $t$ when the particle moves from $x = A$ to $x = \frac{A}{\sqrt{2}}$.

  • At $t=0$, $x = A \cos(0) = A$. This is the starting point.
  • We need to find the time $t$ when $x = \frac{A}{\sqrt{2}}$. $\frac{A}{\sqrt{2}} = A \cos(\omega t)$ $\frac{1}{\sqrt{2}} = \cos(\omega t)$

For the first time this occurs after $t=0$, the angle $\omega t$ must be $\frac{\pi}{4}$ radians.

Final Time Calculation

Now, we solve for $t$: $\omega t = \frac{\pi}{4}$ Substitute the value of $\omega = \frac{2\pi}{5}$: $\left(\frac{2\pi}{5}\right) t = \frac{\pi}{4}$ $t = \frac{\pi}{4} \times \frac{5}{2\pi}$ $t = \frac{5}{8}$ sec.

Therefore, the time required to move from $x = A$ to $x = \frac{A}{\sqrt{2}}$ is $\frac{5}{8}$ sec.

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Similar Questions

  1. The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A\sin\omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T/2\beta$. The value of $\beta$ is ________.
  2. A simple pendulum has a bob with mass $m$ and charge $q$. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is _________. 
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  10. Match List - I with List - II.
    List - IList - II
    A. $\sin^{2} \omega t$I. Periodic with time period $T=\frac{\pi}{\omega}$ but not simple harmonic motion (SHM)
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    Choose the correct answer from the options given below :

Important Questions from Oscillations and Waves

  1. The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A\sin\omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T/2\beta$. The value of $\beta$ is ________.
  2. A simple pendulum has a bob with mass $m$ and charge $q$. The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is _________. 
    (g: acceleration due to gravity)

  3. In an open organ pipe $\nu_3$ and $\nu_6$ are $3^{\text{rd}}$ and $6^{\text{th}}$ harmonic frequencies, respectively. If $\nu_6 - \nu_3 = 2200 \text{ Hz}$ then length of the pipe is _________ mm.
    (Take velocity of sound in air is $330 \text{ m/s}$.)

  4. Using a simple pendulum experiment g is determind by measuring its time period T. Which of the following plots represent the correct relation between the pendulum length L and time period T ?
  5. A cylindrical block of mass $M$ and area of cross section $A$ is floating in a liquid of density $\rho$ and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is ______.
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