Home
Class 12
PHYSICS
A particle of mass m is acted upon by a ...

A particle of mass m is acted upon by a force `F= t^2-kx` . Initially , the particle is at rest at the origin. Then

A

its displacement will be in simple harmonic

B

its velocity will be in simple harmonic

C

its acceleration will be in simple harmonic

D

None of the above

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the force acting on the particle and determine whether its motion can be classified as simple harmonic motion (SHM). ### Step-by-Step Solution: 1. **Understanding the Force**: The force acting on the particle is given by: \[ F = t^2 - kx \] where \( t \) is time, \( k \) is a constant, and \( x \) is the displacement of the particle. 2. **Condition for Simple Harmonic Motion**: For a particle to exhibit simple harmonic motion, the restoring force must be proportional to the negative of the displacement: \[ F = -kx \] This means that the force should only depend on the displacement \( x \) and not on time \( t \). 3. **Analyzing the Given Force**: In the given force \( F = t^2 - kx \), we can see that there is a term \( t^2 \) which depends on time. This indicates that the force is not solely dependent on the displacement \( x \). 4. **Conclusion on Motion Type**: Since the force has a time-dependent term \( t^2 \), it cannot satisfy the condition for SHM. Therefore, the particle will not undergo simple harmonic motion. 5. **Evaluating the Statements**: - **Statement A**: The displacement will be in SHM. **(False)** - **Statement B**: The velocity will be in SHM. **(False)** - **Statement C**: The acceleration will be in SHM. **(False)** - **Statement D**: None of the above. **(True)** Thus, the correct answer is that none of the statements are correct. ### Final Answer: **D: None of the above.**
Promotional Banner

Similar Questions

Explore conceptually related problems

A particle of mass m is acted upon by a force F given by the emprical law F=(R)/(t^2)upsilon(t) If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot :

A particle of mass m is acted on by two forces of equal magnitude F maintaining their orientation relative to the velocity v as shown in Fig. The momentum of the particle

Passage XII)_ A particle of mass m is constrained to move along x-axis. A force F acts on the particle. F always points toward the position labelled E. For example, when the particle is to the left of E,Fpoints to the right. The magnitude of F is consant except at point E where it is zero. The systm 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. Velocity-time graph of the particle is

Passage XII)_ A particle of mass m is constrained to move along x-axis. A force F acts on the particle. F always points toward the position labelled E. For example, when the particle is to the left of E,Fpoints to the right. The magnitude of F is consant except at point E where it is zero. The systm 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. What it is the time period of periodic motion of particle.

A stationary particle of mass m=1.5kg is acted upon by a variable force. The variation of force with respect to displacement is plotted in figure. a. Calculate the velocity acquired by the particle after getting displaced through 6m . b. What is the maximum speed attained by the particle and at what time is it attained?

A block of mass m at rest is acted upon by a force F for a time t. The kinetic energy of block after time t is

A particle is acted upon by a force given by F=(12t-3t^(2)) N, where is in seconds. Find the change in momenum of that particle from t=1 to t=3 sec.

A particle is acted upon by a force F which varies with position x is shown in figure .If the particle at x = 0 kinetic energy of 25J then the kinetic energy of the particle at x = 16 m is

A particle of mass 2kg is moving in free space with velocity vecv_0=(2hati-3hatj+hatk)m//s is acted upon by force vecF=(2hati+hatj-2hatk)N . Find velocity vector of the particle 3s after the force starts acting.

A particle moving along the x axis is acted upon by a single force F = F_0 e^(-kx) , where F_0 and k are constants. The particle is released from rest at x = 0. It will attain a maximum kinetic energy of :