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Physics Question on simple harmonic motion

The SHMSHM of a particle is given by x(t)=5cos(2πt+π4)x\left(t\right)=5\,cos\left(2\pi t+\frac{\pi}{4}\right) (in MKSMKS units). Calculate the displacement and the magnitude of acceleration of the particle at t=1.5st =1.5\,s .

A

3.0m-3.0\, m , 100m/s2100 \,m/s^2

B

+2.54m+2.54\,m , 200m/s2200 \,m/s^2

C

3.54m-3.54\,m , 140m/s2140 \,m/s^2

D

+3.55m+3.55\, m , 120m/s2120\, m/s^2

Answer

3.54m-3.54\,m , 140m/s2140 \,m/s^2

Explanation

Solution

Displacement x(t)=5\,cos(2π×32+π4)x \left(t\right)=5\backslash,cos \left(2\pi\times\frac{3}{2}+\frac{\pi}{4}\right) 5cos[13π4]=3.5m5\,cos \left[\frac{13 \pi}{4}\right]=-3.5\,m y=5cos(2πt+π4)y=5\,cos \left(2\pi t+\frac{\pi}{4}\right) V=10πsin(2πt+π4)\therefore V=-10\pi\,sin \left(2\pi t+\frac{\pi}{4}\right) \therefore acceleration =20π2cos(2πt+π4)= -20\pi^{2} cos \left(2\pi t+\frac{\pi}{4}\right) =20π2cos(2π×32+π4)=-20\pi^{2}\, cos \left(2\pi\times\frac{3}{2}+\frac{\pi}{4}\right) =20π2cos13π4=20\pi^{2}\,cos \frac{13\pi}{4} =140m/s2=140 m/ s^{2}

The smallest distance between an object's original and final positions is called displacement. It is a change in both the object's location and the direction of motion.

Since the starting and ending locations are always connected in order to determine displacement, the route of displacement is always straight.

The letter 'S' is used to denote it.

Displacement may be zero, positive, or negative.

Given that it has both a magnitude and a direction, it is a vector quantity.

The displacement of an item is measured using the displacement formula. As stated, the displacement formula is S = Sf - Si. In this case, S stands for displacement, Sf for the object's final location, and Si for its starting position.

The alteration of an object's location is known as displacement. The word "displacement" denotes a movement or displacement of the thing from its original position.

  • Displacement is concerned with linear motion.
  • It is the shift in an object's position between its original location and its ultimate position.
  • It is a vector quantity since it has both a direction and a magnitude.
  • Its value might be zero, negative, or positive.

The rate at which velocity changes in relation to time is known as acceleration.

  • Only when an item changes its speed, direction, or both can it accelerate.
  • The velocity may alter due to a change in direction of travel, an increase or reduction in speed.
  • Given that it has both a magnitude and a direction, it is a vector quantity.
  • The acceleration unit in the SI is m/s2.

The rate at which an object's velocity alters over time is known as its acceleration. A body is said to have accelerated if it modifies its direction, speed, or both.

  • 'A' stands for acceleration, and the SI unit for acceleration is m/s2.
  • Acceleration is a vector quantity since it has both magnitude and direction.
  • It is calculated by dividing the change in velocity by the time period.
  • It is the position's second derivative with regard to time.
  • It is the velocity's first derivative with regard to time.
  • An object accelerates when its velocity and acceleration are in the same direction.
  • The item slows down when the velocity and acceleration are in the opposing directions.

According to the three equations of motion, Acceleration Formula is given as follows:

  1. v = u + at
  2. v2 = u2 + 2as
  3. s = ut + ½ at2
  • u: Initial Velocity
  • v: Final Velocity
  • t: Time Taken
  • a: Acceleration
  • s: Distance/Displacement