OSCILLATIONS AND WAVES

11 ) Free Physics Questions Online Mock Test Series By Cracktest

A particle executing S.H.M. has an amplitude A and periodic time T. The minimum time required by the particle to get displaced by (A / √2) from its equilibrium position is …….. s.

T/8

T/16

T/4

View Solution

Answer :

y = A sin (ωt + ɸ)

(A / √2) = A sin (ωt) ----- ɸ = 0

∴ sin ωt = (1 / √2)

ωt = sin^–1 (1 / √2) = (π/4)

∴ (2π / T) × t = (π/4)

∴ t = (T / 8)

12 )

If a body having mass M is suspended from the free ends of two springs A and B, their periodic time are found to be T₁ and T₂ respectively. If both these springs are now connected in series and if the same mass is suspended from the free end, then the periodic time is found to be T. Therefore ………..

Select correct option from the following options

(A) T = T₁ + T₂ (B) (1/T) = (1/T₁) + (1/T₂)

View Solution

Answer :

T₁ = 2π √(M / K₁), T₂ = 2π (M / K₂) (1)

when both are connected in series, spring constant = K_s.

hence

(1 / K_s) = (1 / K₁) + (1 / K₂) (2)

& T = 2π √(M / Ks) ---- mass remains same (3)

from (1),

T₁² + T₂² = 4π² {(M / K₁) + (M / K₂)}

from (3)

T² = 4π² (M / K_s)

putting value of Ks from (2)

∴ T² = 4π² [M / (1 / {(1/K₁) + (1 / K₂)})]

T² = 4π² × M {(1 / K₁) + (1 / K₂)}

T² = 4π² × [(M / K₁) + (M / K₂)]

hence T² = T₁² + T₂²

13 )

The displacement of a S.H.O. is given by the equation x = A cos {ωt + (π/8)}. At what time will it attain maximum velocity ?

(3Ï€ / 8Ï‰)

(8Ï€ / 3Ï‰)

(3Ï€ / 16Ï‰)

(Ï€ / 16Ï€)

View Solution

Answer :

x = A cos {ωt + (π/8)}

(dx / dt) = V = A (– ω) sin {ωt + (π/8)}

Velocity will be maximum,

if sin {ωt + (π/8)} = 1

i.e. ωt + (π/8) = sin^–1(1)

ωt + (π/8) = (π/2)

ωt = {3π / 8)

t = (3π / 8ω)

14 )

At what position will the potential energy of a S.H.O. become equal to one third its kinetic energy ?

Â± (A / 2)

Â± (A / âˆš2)

Â± (A / âˆš3)

Â± âˆš3 A

View Solution

Answer :

given: PE = (1/ 3) KE

(1/2) Ky² = (1/3) × (1/2) K (A² – y²)

∴ y² = {(A² – y²) / 3}

∴ 3y² = A² – y²

4y² = A²

y² = (A² /4) i.e. y = ± (A / 2)

15 )

Three identical springs are shown in figure. When a 4 kg mass is suspended from spring A, its length increases by 1cm. Now if a 6 kg mass is suspended from the free end of spring C, then increase in its length is ……..cm.

1.5

3.0

4.5

6.0

View Solution

Answer :

F = Kx i.e. mg = Kx (1)

spring A, m₁ = 4 kg, x₁ = 1 cm, spring constant = K₁ = K

combination of spring B & C, m₂ = 6kg. let extension be x₂ & spring

constant = K₂ = {(K ∙ K) / (K + K)} = (K² / 2K) = (K / 2)

∴ from (1)

{(m₁ g) / (m₂ g)} = {(K₁ x₁) / (K₂ x₂)}

∴ (4/6) = {K / (K/2)} {1 / x₂}

∴ (2/3) = 2 (1 / x₂)

∴ x₂ = 3 cm

Topic Categories