Q: सरल आवर्ती गति करते किसी कण के विस्थापन और त्वरण के बीच कलान्तर होता है:

(d) विस्थापन समीकरण स० आ० गति में,y=Asin(ωt+ϕ)∴ वेग, v=dtdy=Aωcos(ωt+ϕ)त्वरण, a=dtdv या, a=−Aω2sin(ωt+ϕ) ∴a=Aω2sin(ω+ϕ)अतः विस्थापन तथा त्वरण के बीच कालांतर π है।

Q: A point mass oscillates along the X-axis according to the relation x=x0cos(ωt−π/4). If acceleration of the particle is written as a=a0cos(ωt+δ), then

x=x0cos(ωt−π/4)v=dtdx=−x0ωsin(ωt−π/4)a=dtdv=−x0ω2cos(ωt−π/4) =+x0ω2cos[π+(ωt−π/4)] =x0ω2cos(ωt+3π/4)Comparing it with given equation, a=a0cos(ωt+δ) ⇒a0=x0ω2 and δ=43π

Question 1

Answer the following questions :(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:T=2πkm​​. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angle of oscillation a more involved analysis shows that T is greater than  2πgl​​. Think of a qualitative argument to appreciate this result.(c) A man with a wristwatch on his hand falls from the top of a tower. Does thewatch give correct time during the free fall ?(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity ?

Accepted Answer

(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m cancels out in denoator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:F =–mgsinθ where,F= Restoring forcem= Mass of the bobg= Acceleration due to gravityθ= Angle of displacementFor small θ,sinθ≃ θFor large θ, sinθ is greater than θ.This decreases the effective value of g.Hence, the time period increases as:T=2πgl​​(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.(d) Gravity disappears for a object under free fall, so frequency is zero.

Question 2

सरल आवर्ती गति करते किसी कण के विस्थापन और त्वरण के बीच कलान्तर होता है:

Accepted Answer

(d) विस्थापन समीकरण स० आ० गति में,y=Asin(ωt+ϕ)∴ वेग, v=dtdy​=Aωcos(ωt+ϕ)त्वरण, a=dtdv​ या, a=−Aω2sin(ωt+ϕ)     ∴a=Aω2sin(ω+ϕ)अतः विस्थापन तथा त्वरण के बीच कालांतर π है।

Question 3

A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1​(t)=Asinsinωt and x2​(t)=Asinsin(ωt+32π​). Adding a third sinusoidal displacement x3​(t)=Bsinsin(ωt+ϕ) brings the mass to a complete rest. The values of B and ϕ

Accepted Answer

So B=A,ϕ=240∘=34π​

Question 4

Y=Asin(ωt+ϕ0​)  is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is Y=2A​ and it is moving along negative x-direction. Then the initial phase angle ϕ0​ will be :

Accepted Answer

Draw phasor :At B particle is going toward mean position⇒ negative x-directionϕ=65π​

Question 5

A point mass oscillates along the X-axis according to the relation x=x0​cos(ωt−π/4). If acceleration of the particle is written as a=a0​cos(ωt+δ), then

Accepted Answer

x=x0​cos(ωt−π/4)v=dtdx​=−x0​ωsin(ωt−π/4)a=dtdv​=−x0​ω2cos(ωt−π/4) =+x0​ω2cos[π+(ωt−π/4)] =x0​ω2cos(ωt+3π/4)Comparing it with given equation, a=a0​cos(ωt+δ) ⇒a0​=x0​ω2 and δ=43π​

Filters

Filters

class

subject

topic

concept

difficulty

Solutions for Analysing shm through phasor diagram Important Questions

Found the solution, but did not understand the concept?

Ask my first question