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A 1kg block is executing simple harmonic...

A `1kg` block is executing simple harmonic motion of amplitude `0.1m` on a smooth horizontal surface under the restoring force of a spring of spring constant `100 N//m`. A block of mass `3 kg` is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together. Find the frequency and the amplitude of the motion.

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AI Generated Solution

To solve the problem step by step, we will first find the frequency of the original block and then calculate the new frequency and amplitude after the second block is placed on it. ### Step 1: Calculate the Frequency of the Original Block The formula for the frequency \( f_0 \) of a mass-spring system in simple harmonic motion is given by: \[ f_0 = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \] ...
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A 1kg block is eecuting simpe hrmonic motion of ampliltude 0.1 m m on a smooth horizontal surface under the restoring force of a spring of sprign constant 100Nm^-1 . A block of mass 3 kg is gently placed on it at the instant it pases through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.

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Knowledge Check

  • A block of mass m moving with velocity v_(0) on a smooth horizontal surface hits the spring of constant k as shown. Two maximum compression in spring is

    A
    `sqrt((2m)/(k))v_(0)`
    B
    `sqrt((m)/(k)).v_(0)`
    C
    `sqrt((m)/(k)).v_(0)`
    D
    `(m)/(2k).v_(0)`

  • A block of mass 1 kg kept over a smooth surface is given velocity 2 m//s towards a spring of spring constant 1 N//m at a distance of 10 m . Find after what time block will be passing through P again .

    A
    `(20+2 pi)sec`
    B
    `10 sec`
    C
    `(10 + 2pi) sec`
    D
    `(10 + pi) sec`

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