Home
Class 11
PHYSICS
Assertion : In x=5-4"sinomegat , motion ...

Assertion : In x=5-4`"sinomegat` , motion of body is SHM about the mean position x=5
Reason Amplitude of oscillation is 9.

A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

B

If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.

C

If Assertion is true but Reason is false.

D

If Assertion is false but Reason is true.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question step by step, let's analyze the assertion and reason provided. ### Step 1: Understand the Assertion The assertion states that the equation \( x = 5 - 4 \sin(\omega t) \) describes simple harmonic motion (SHM) about the mean position \( x = 5 \). ### Step 2: Identify the Mean Position In SHM, the mean position is where the displacement is zero. From the equation \( x = 5 - 4 \sin(\omega t) \), we can see that when \( \sin(\omega t) = 0 \), \( x \) reaches its mean position: \[ x = 5 - 4 \cdot 0 = 5 \] Thus, the mean position is indeed \( x = 5 \). ### Step 3: Determine the Amplitude The amplitude of SHM is defined as the maximum displacement from the mean position. In the equation \( x = 5 - 4 \sin(\omega t) \), the term \( -4 \) indicates that the maximum displacement from the mean position (5) is 4 units. Therefore, the amplitude is: \[ \text{Amplitude} = 4 \text{ units} \] ### Step 4: Analyze the Reason The reason states that the amplitude of oscillation is 9 units. However, we have calculated that the amplitude is actually 4 units, not 9. Therefore, the reason is incorrect. ### Step 5: Conclusion Based on our analysis: - The assertion is true: The motion described is indeed SHM about the mean position \( x = 5 \). - The reason is false: The amplitude is not 9 units; it is 4 units. Thus, the correct answer is that the assertion is true, but the reason is false. ### Final Answer Assertion is true, Reason is false. ---

To solve the question step by step, let's analyze the assertion and reason provided. ### Step 1: Understand the Assertion The assertion states that the equation \( x = 5 - 4 \sin(\omega t) \) describes simple harmonic motion (SHM) about the mean position \( x = 5 \). ### Step 2: Identify the Mean Position In SHM, the mean position is where the displacement is zero. From the equation \( x = 5 - 4 \sin(\omega t) \), we can see that when \( \sin(\omega t) = 0 \), \( x \) reaches its mean position: \[ ...
Promotional Banner

Similar Questions

Explore conceptually related problems

Assertion : In x=3+4 "cos" omegat , amplitude of oscillation is 4 units. Reason : Mean position is at x=3.

Potential energy of a particle at mean position is 4 J and at extreme position is 20 J. Given that amplitude of oscillation is A. Match the following two columns

The force on a body executing SHM is 4 N when the displacement from mean position is 2 cm. If amplitude of oscillation is 10 cm, then the maximum kinetic energy associated with the SHM will be

The kinetic energy of a particle, executing S.H.M. is 16 J when it is in its mean position. IF the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 Kg, the time period of its oscillations is

A particle of mass 0.2 kg exectites SHM under a force of F = - 20x N. If speed of particle at mean position is 12 m/s then the amplitude of oscillations is

A particle of mass 0.1 kg executes SHM under a for F=(-10x) N. Speed of particle at mean position 6m//s . Then amplitude of oscillations is

The SHM of a particle is given by the equation x=2 sin omega t + 4 cos omega t . Its amplitude of oscillation is

Potential energy of a particle in SHM along x - axis is gives by U = 10 + (x - 2)^(2) Here, U is in joule and x in metre. Total mechanical energy of the particle is 26 J . Mass of the particle is 2kg . Find (a) angular frequency of SHM, (b) potential energy and kinetic energy at mean position and extreme position, (c ) amplitude of oscillation, (d) x - coordinates between which particle oscillates.

A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 4 xx 10^(-3) m. The least period of these oscillations, so that the object is not detached from the platform is pi/(5x) second. Find value of x. Take g=10m/ s^(2)

Displacement-time equation of a particle executing SHM is x=4sin( omega t) + 3sin(omegat+pi//3) Here, x is in cm and t ini sec. The amplitude of oscillation of the particle is approximately.