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A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand.
(a) What is the amplitude of oscillation?
(b) Find the frequency of oscillation?

Answers (1)

(a) As no deforming force acts on the spring when mass ‘m’ is supported by hand extension in the spring. Let m reach its new position at displacement = x unit, from the previous one, then,

P.E. of the spring or mass = gravitational P.E. lost by man

P.E. = mgx

But due to spring,

$\frac{1}{2} kx^{2}k = \omega ^{2}A$

Thus, $x = \frac{2mg}{k}$

When extension is $x_{0}$ , the spring will be at the mean position by the block

$F=kx_{0}$

F = mg

Thus, $x_{0}=\frac{mg}{k}$ ……. (ii)

Now, from (i) & (ii)

$x=2\left (\frac{mg}{k} \right )$

= $2x_{0} = 4cm$

Thus, $x_{0} = 2cm$

$x - x_{0} = 4-2 = 2cm.$

Thus, from the mean position, the amplitude of the oscillator is maximum.

(b) Now, we know that time period (T) does not depend on amplitude,

$T = 2\pi \sqrt{\frac{m}{k}}$

From (i)

$\frac{2mg}{k} = x$

$\frac{m}{k} = \frac{x}{2g}$

$= \frac{4 \times 10^{-2}}{2\times 9.8}$

Or, $\frac{k}{m}= \frac{2\times 9.8}{4\times 10^{-2}}$

Now, $v = \frac{1}{2}\pi \sqrt{\frac{k}{m}}$

$= \frac{1}{2}\times 3.14\sqrt{\frac{2 \times 9.8}{4 \times 10^{-2}} }$

$= \frac{10 \times 2.21}{6.28}$

$= 3.52 Hz$

Since the total extension in the spring is 4 cm when released & amplitude is 2 cm, the oscillator will not rise above 4cm, & hence, it will oscillate below the released position.

Posted by

infoexpert24

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