The displacement versus time graph of SHM is as shown in figure :
Which of the following is its acceleration versus time graph ?
Equation of y-t graph is : y = A sin (ωt) differentiating v = A ω cos (ωt) and a= - Aω² sin (ωt) . So the graph is a function of - sin(ωt).
A particle is in linear SHM between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction, find the signs of velocity, acceleration and force on the particle when it is 4 cm from B going towards A :
For a particle in simple harmonic motion (SHM), the force and acceleration are always directed towards the equilibrium position (center) and are opposite to the displacement.
1. Displacement: Since the particle is 4 cm from \( B \) and moving towards \( A \), it is located closer to \( B \) on the negative side of the equilibrium point. This means the displacement is in the negative direction.
2. Velocity: Since the particle is moving towards \( A \) (the negative direction), its velocity is negative.
3. Acceleration and Force: In SHM, acceleration and force are always directed towards the equilibrium point. Since the particle is on the positive side of the equilibrium (closer to \( B \)), the acceleration and force are in the negative direction.
Conclusion:
The signs of velocity, acceleration, and force are all negative (–, –, –).
The equation of SHM of a particle is a + 4π²x = 0 where a is instantaneous linear acceleration at displacement x. The frequency of motion is :
No solution provided for this question.
The velocities of a body executing SHM are v 1 and v 2 when the displacements are x1 and x2 respectively. The period is :
A particle executes an undamped SHM of time period T. Then the time period with which the potential energy changes is :
If equation of motion in shm is x = A sin ωt then eqn. of potential energy = ½k x²= ½kA² sin²ωt
\[ U = \frac{1}{2}K A^{2}\left( \frac{1- sin(2\omega t)}{2} \right) \]
Frequency becomes double so time period becomes half.
Two particles execute SHM of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions, each time their displacement is half of their amplitude. The phase difference between them is :
The equation of a simple harmonic motion is given by, y = 3sin π/2 (50t – Φ), where y is in metres and t is in sec, the maximum particle velocity in ms–¹ is :
y = 3sin π/2 (50t - Φ)
\[ \frac{\partial y}{\partial t}= 3 \times \frac{\Pi}{2}\times 50 cos(π/2 (50t - Φ))\]
Maximum speed = 75 π
A periodic time of a body executing simple harmonic motion is 3s. After how much interval from time t = 0, its displacement from mean position will be half of its amplitude ?
\[ x= A sin\left( \omega t \right) \]
\[ \frac{A}{2}= A sin\left( \omega t \right) \]
\[ \frac{1}{2}= sin\left( \omega t \right) \]
\[ \frac{\Pi}{6}= \omega t =\frac{2\Pi}{T}t = \frac{2\Pi}{3}t \]
\[ t= \frac{1}{4}s \]
Displacement-time graph of a particle executing SHM is as shown below :-
The corresponding force-time graph of the particle can be :-
Equation of displacement x = A sin (ωt) so, acceleration a= -Aω sin(ωt) force will have same nature so,
The equation of motion of a particle of mass 1 gm is d²x/dt² + π²x = 0 where x is displacement (in m) from mean position. The frequency of oscillation is (in Hz) :
