For `SHM` to take place force acting on the body should be proportional to `-x` or `F =- kx`. If `A` be the amplitude then energy of oscillation is `1//2KA^(2)`.
Force acting on a block is `F = (-4x +8)`. Here `F` is in newton and `x` in the position of block on x-axis in meters
For `SHM` to take place force acting on the body should be proportional to `-x` or `F =- kx`. If `A` be the amplitude then energy of oscillation is `1//2KA^(2)`.
Force acting on a block is `F = (-4x +8)`. Here `F` is in newton and `x` in the position of block on x-axis in meters
Force acting on a block is `F = (-4x +8)`. Here `F` is in newton and `x` in the position of block on x-axis in meters
A
Motion of the block is periodic but not simple harmonic
B
Motion of the block is not period
C
Motion of the block is simple harmonic about the origin, `x =0`
D
Motion of the block is simple harmonic about `x =2m`
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem, we need to analyze the given force equation and determine the characteristics of the motion of the block.
### Step 1: Identify the given force equation
The force acting on the block is given by:
\[ F = -4x + 8 \]
### Step 2: Rearrange the force equation
We can rewrite the force equation as:
\[ F = -4(x - 2) \]
This shows that the force can be expressed in the form of a restoring force, where the equilibrium position (mean position) is at \( x = 2 \).
### Step 3: Identify the spring constant (k)
From the rearranged equation, we can see that the coefficient of \( x \) is \(-4\). Therefore, we can identify:
\[ k = 4 \]
### Step 4: Determine the mean position (x₀)
From the equation \( F = -k(x - x₀) \), we can see that the mean position \( x₀ \) is:
\[ x₀ = 2 \]
### Step 5: Determine the amplitude (A)
In simple harmonic motion (SHM), the amplitude is the maximum displacement from the mean position. Since the force equation indicates a linear restoring force, the motion can be periodic. The amplitude can be considered as the distance from the mean position to the maximum displacement. If we assume the maximum displacement is \( A \), then:
\[ A = 4 \] (as inferred from the force equation)
### Step 6: Determine the type of motion
Since the force is not purely proportional to \(-x\) (due to the constant term +8), the motion is not simple harmonic motion. The motion is periodic but not simple harmonic because the restoring force does not follow the form \( F = -kx \) around the origin.
### Conclusion
The motion of the block is periodic but not simple harmonic due to the presence of the constant force term.
### Final Answer
The motion of the block is periodic but not simple harmonic.
---
To solve the problem, we need to analyze the given force equation and determine the characteristics of the motion of the block.
### Step 1: Identify the given force equation
The force acting on the block is given by:
\[ F = -4x + 8 \]
### Step 2: Rearrange the force equation
We can rewrite the force equation as:
...
Topper's Solved these Questions
Similar Questions
Explore conceptually related problems
Passage IX) For SHM to take place force acting on the body should be proportional to -x or F = -kx. If A be the amplitude then energy of oscillation is 1/2 k A^(2) Force acting on a block is F=(-4x +8) . Here F is in Newton and x the position of block on x-axis in meter
For SHM to take place force acting on the body should be proportional to -x or F =- kx . If A be the amplitude then energy of oscillation is 1//2KA^(2) . The amplitude of oscillation is
For SHM to take place force acting on the body should be proportional to -x or F =- kx . If A be the amplitude then enegry of oscillation is 1//2KA^(2) . If enegry of oscillation is 18J , between what points does the block will oscillate?
Passage IX) For SHM to take place force acting on the body should be proportional to -x or F = -kx. If A be the amplitude then energy of oscillation is 1/2 k A^(2) If energy of osciallation is 18 J, between what points the block will oscillate?
F - x equation of a body of mass 2kg in SHM is F + 8x = 0 Here, F is in newton and x in meter.Find time period of oscillations.
The spring force is given by F=-kx , here k is a constant and x is the deformation of spring. The F-x graph is
F-x equation of a body in SHM is F + 4x=0 Here, F is in newton and x in meter. Mass of the body is 1 kg . Find time period of oscillations.
The damping force acting on an oscillating body is proportional to its velocity i.e. F = Kv. What is the dimensional formula for K?
A force F = - kx^(3) is acting on a block moving along x-axis. Here, k is a positive constant. Work done by this force is
A force F=-4x+8 (in N) is acting on a block where x is the position of the block in metres. The energy of oscillation is 32 J. The block oscillates between two points, out of which the value of position of one point (in metres) is an integer from 0 to 9. Find it.
Knowledge Check
Passage IX) For SHM to take place force acting on the body should be proportional to -x or F = -kx. If A be the amplitude then energy of oscillation is 1/2 k A^(2) Force acting on a block is F=(-4x +8) . Here F is in Newton and x the position of block on x-axis in meter
Passage IX) For SHM to take place force acting on the body should be proportional to -x or F = -kx. If A be the amplitude then energy of oscillation is 1/2 k A^(2) Force acting on a block is F=(-4x +8) . Here F is in Newton and x the position of block on x-axis in meter
A
motion of the block is periodic but not simple harmonic
B
Motion of the block is not periodic
C
motion of the block is simple harmonic about origin, x=0
D
motion of the block is simple harmonic about x=2m.
For SHM to take place force acting on the body should be proportional to -x or F =- kx . If A be the amplitude then energy of oscillation is 1//2KA^(2) . The amplitude of oscillation is
For SHM to take place force acting on the body should be proportional to -x or F =- kx . If A be the amplitude then energy of oscillation is 1//2KA^(2) . The amplitude of oscillation is
A
`4cm`
B
`2cm`
C
`1cm`s
D
`3cm`
For SHM to take place force acting on the body should be proportional to -x or F =- kx . If A be the amplitude then enegry of oscillation is 1//2KA^(2) . If enegry of oscillation is 18J , between what points does the block will oscillate?
For SHM to take place force acting on the body should be proportional to -x or F =- kx . If A be the amplitude then enegry of oscillation is 1//2KA^(2) . If enegry of oscillation is 18J , between what points does the block will oscillate?
A
between `x =0` nad `x =4m`
B
between `x =-1m` and `x =5 m`
C
between `x =-2` and `x =6m`
D
between `x =1m` and `x =3m`