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1. ⇒  (MHT CET 2023 13th May Morning Shift )

A body of mass 0.04   kg executes simple harmonic motion (SHM) about x = 0 under the influence of force F as shown in graph. The period of

MHT CET 2023 13th May Morning Shift Physics - Simple Harmonic Motion Question 1 English

A. 2 π s

B. 0.2 π s

C. π d

D. π 2   s

Correct Option is (B)

k = 400   N / m m = 0.04

From the graph,

K = F x = 8 2 = 4

From T = 2 π M K ,

we get

T = 2 π 0.04 4 = 0.2 π s

2. ⇒  (MHT CET 2023 10th May Evening Shift )

A simple harmonic progressive wave is represented by y = A sin ( 100 π t + 3 x ) . The distance between two points on the wave at a phase difference of π 3 radian is

A. π 8   m

B. π 9   m

C. π 6   m

D. π 3   m

Correct Option is (B)

Equation of the given harmonic progressive wave y = A sin ( 100 π t + 3 ) ...... (i)

General equation of a harmonic wave

y = A sin ( wt + kx ) ..... (ii)

From equations (i) and (ii), ω = 100 π , k = 3

But, k = 2 π λ 3 = 2 π λ

λ = 2 π 3

We also know,

Path difference Δ x = λ 2 π × Phase difference Δ ϕ

Δ x = 2 π 3 × 2 π × π 3 . (  Given  Δ ϕ = π 3 ) = π 9   m

3. ⇒  (MHT CET 2023 9th May Evening Shift )

Two S.H.Ms. are represented by equations y 1 = 0.1 sin ( 100 π t + π 3 ) and y 2 = 0.1 cos ( 100 π t ) The phase difference between the speeds of the two particles is

A. π 3

B. π 6

C. + π 6

D. π 3

Correct Option is (B)

Given: y 1 = 0.1 sin ( 100 π t + π 3 ) and

y 2 = 0.1 cos ( 100 π t ) v 1 = dy 1 dt = 0.1 × 100 π cos ( 100 π t + π 3 ) v 2 = dy 2 dt = 0.1 × 100 π sin ( 100 π t ) = ( 0.1 × 100 π ) cos ( 100 π t + π 2 )

Phase difference of velocity of first particle with respect to the velocity of second particle at t = 0 is

Δ ϕ = ϕ 1 ϕ 2 = π 3 π 2 = π 6

4. ⇒  (MHT CET 2023 9th May Morning Shift )

The maximum velocity of a particle performing S.H.M. is ' V '. If the periodic time is made ( 1 3 ) d and the amplitude is doubled, then the new maximum velocity of the particle will be

A. V 6

B. 3   V 2

C. 3   V

D. 6   V

Correct Option is (D)

Given T = 1 3   T and A = 2   A

ω = 2 π T = 2 π ( 1 3 T ) = 6 π T = 3 ω

 The new maximum velocity  V = A ω = ( 2   A ) × ( 3 ω ) = 6   A ω = 6   V

5. ⇒  (MHT CET 2021 21th September Evening Shift )

A particle is performing S.H.M. with maximum velocity ' v '. If the amplitude is tripled and periodic time is doubled then maximum velocity will be

A. 1.5   v

B. 3   v

C. 2   v

D. v

Correct Option is (A)

V = V max = ω A = 2 π T   .   A

If A is tripled and T is doubled, V max will become 3 2 times or 1.5 times.