OSCILLATIONS%20AND%20WAVES

1 ) Free Physics Questions Online Mock Test Series By Cracktest

If the equation for a particle performing S.H.M. is given by y = sin 2t + √3 cos 2t, its periodic time will be …….. s.

Ï€

2Ï€

4Ï€

View Solution

Answer :

y = sin 2t + √3 + cos 2t

∴ y = 2[(1/2) sin 2t + (√3 / 2) cos 2t]

= 2[cos ɸ sin 2t + sin ɸ cos 2t]

= 2 [sin (2t + ɸ)

comparing with y = A sin (ωt + ɸ) we get

ω = 2

i.e. (2π / T) = 2

∴ T = π sec

2 )

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to T. Now, if the lift moves along the vertically upward direction with an acceleration of (g/3), then the periodic time of the lift will now be……………

âˆš3T

âˆš(3/2)T

(T/3)

(T / âˆš3)

View Solution

Answer :

T = 2π √(ℓ/g)

in stationary lift, time periods is T. When lift moves up with acceleration

(g/3), effective acceleration is

g' = g + (g/3) = (4/3)g

∴ New time period = T' = 2π √(ℓ / g') = 2π √{ℓ / (4/3)g} = √{(3/4) (ℓ/g)} × 2π

T' = √(3/2) × 2π √(ℓ/g)

T' = √(3/2)T

3 )

If the equation for displacement of two particles executing S.H.M. is given by y₁ = 2 sin (10t + θ) and y₂ = 3 cos 10t respectively, then the phase difference between the velocity of two particles will be ……….

Î¸

Î¸ â€“ (Ï€/2)

â€“ Î¸

Î¸ + (Ï€/2)

View Solution

Answer :

y₁ = 2 sin (10t + θ)

v₁ = (dy₁ / at) = 2 cos (10t + θ) × 10 = 20 cos (10t + θ)

y₂ = 3 cos 10t

v₂ = (dy₂ / at) = 30 (– sin 10t)

= 30 cos {10t + (π/2)}

phase difference = (10t + θ) – {10t + (π/2)}

= θ – (π/2)

4 )

When a body having mass m is suspended from the free end of two springs suspended from a rigid support, as shown in figure, its periodic time of oscillation is T. If only one of the two springs are used, then the periodic time would be ………

(T / âˆš2)

(T/2)

âˆš2 T

View Solution

Answer :

for series combination, (1 / k_s) = (1 / k₁) + (1 / k₂)

here k₁ = k₂ = K (say) then (1 / k_s) = (1/k) + (1/k) = (2/k)

∴ k_s = (k/2)

when single spring is used then spring constant is K.

as T = 2π √(m/k)

T ∝ (1 / √k) i.e. (T₁ / T₂) = √(k₂ / k₁)

here T₁ = T

k₁ = (k / 2)

k₂ = k

∴ (T₁ / T₂) = √{k / (k/2)} = √2

∴ T₂ = (T₁ / √2) = (T / √2)

5 )

If the maximum velocity of two springs (both has same mass) executing S.H.M. and having force constants k₁ and k₂ respectively are same, then the ratio of their amplitudes will be ……..

Select correct option from following options

(A) (k₁ / k₂) (B) (k₂ / k₁)

View Solution

Answer :

For maximum velocity A₁ ω₁ = A₂ ω₂

∴ (A₁ / A₂) = (ω₂ / ω₁)

as ω = √(k/m), here mass of both springs are equal hence

ω ∝ √k

∴ (A₁ / A₂) = √(k₂ / k₁)

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