Oscillations

Oscillations

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Moderate Questions

Physics of Sound

Oscillations
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  1. A particle is executing a simple harmonic motion. Its maximum acceleration is a and maximum velocity is b.Then its time period of vibration will be :








  1. View Hint View Answer Discuss in Forum

    As, we know, in SHM
    ​Maximum acceleration of the particle, α = Aω2 ​
    Maximum velocity, β = Aω

    ⇒ ω =
    α
    β

    T =
    2π
    		 -
    2πβ
    		 ∵
    2π
    ωαT

    Correct Option: C

    As, we know, in SHM
    ​Maximum acceleration of the particle, α = Aω2 ​
    Maximum velocity, β = Aω

    ⇒ ω =
    α
    β

    T =
    2π
    		 -
    2πβ
    		 ∵
    2π
    ωαT

  1. A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2, respectively. Its time period is​​​








  1. View Hint View Answer Discuss in Forum

    As we know, for particle undergoing SHM,
    V = ω √A² - X²
    V12 = ω2 (A2 - x12)
    V22 = ω2 (A2 - x22)
    Substracting we get,

    V12
    		 + x12 =
    V22
    		 + x22
    ω2 ω2

    ⇒
    V12 - V22
    		 = x22 - x12
    ω2

    ⇒ w =
    V12 - V22
    x22 - x12

    ⇒ T = 2π
    x22 - x12
    V12 - V22

    Correct Option: A

    As we know, for particle undergoing SHM,
    V = ω √A² - X²
    V12 = ω2 (A2 - x12)
    V22 = ω2 (A2 - x22)
    Substracting we get,

    V12
    		 + x12 =
    V22
    		 + x22
    ω2 ω2

    ⇒
    V12 - V22
    		 = x22 - x12
    ω2

    ⇒ w =
    V12 - V22
    x22 - x12

    ⇒ T = 2π
    x22 - x12
    V12 - V22

  1. A spring of force constant k is cut into lengths of ratio 1 : 2 : 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k′′ . Then k' : k′′ is​​​








  1. View Hint View Answer Discuss in Forum

    Let ℓ be the complete length of the spring. ​

    Length when cut in ratio, 1 : 2 : 3 are
    ℓ
    		 ,
    ℓ
    		 and
    ℓ
    632

    Spring constant (k) ∝
    1
    length (ℓ)

    Spring constant for given segments ​k1 = 6k, k2 = 3k and k3 = 2k ​
    When they are connected in series
    1
    		 =
    1
    		 +
    1
    		 +
    1
    k'6k3k2k

    ⇒
    1
    		 =
    6
    k'6k

    ∴ Force constant  k' = k ​
    And when they are connected in parallel ​k" = 6k + 3k + 2k ​
    ⇒ k" = 11k ​
    Then the ratios
    k'
    		 =
    1
    		 i.e. k' : k" = 1 : 11
    k"11

    Correct Option: B

    Let ℓ be the complete length of the spring. ​

    Length when cut in ratio, 1 : 2 : 3 are
    ℓ
    		 ,
    ℓ
    		 and
    ℓ
    632

    Spring constant (k) ∝
    1
    length (ℓ)

    Spring constant for given segments ​k1 = 6k, k2 = 3k and k3 = 2k ​
    When they are connected in series
    1
    		 =
    1
    		 +
    1
    		 +
    1
    k'6k3k2k

    ⇒
    1
    		 =
    6
    k'6k

    ∴ Force constant  k' = k ​
    And when they are connected in parallel ​k" = 6k + 3k + 2k ​
    ⇒ k" = 11k ​
    Then the ratios
    k'
    		 =
    1
    		 i.e. k' : k" = 1 : 11
    k"11

  1. A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is​








  1. View Hint View Answer Discuss in Forum

    Given, Amplitude A = 3 cm ​
    When particle is at x = 2 cm ​
    According to question, magnitude of velocity = acceleration
    ω √A² - x² = x ω2

    √3² - 2² = 2
    2π
    T

    √5 =
    4π
    		 ⇒ T =
    4π
    T√5

    Correct Option: B

    Given, Amplitude A = 3 cm ​
    When particle is at x = 2 cm ​
    According to question, magnitude of velocity = acceleration
    ω √A² - x² = x ω2

    √3² - 2² = 2
    2π
    T

    √5 =
    4π
    		 ⇒ T =
    4π
    T√5