First slide

Simple harmonic motion

Question

A body of mass m is hanging from a wire of length l. It is pulled by a distance $Δ l$ and left. It executes simple harmonic motion. If the young’s modulus of the wire is Y, area of cross-section is A, then the frequency of oscillation is

Easy

A

$f = \frac{1}{2 π} \sqrt{\frac{m l}{Y A}}$
B

$f = \frac{1}{2 π} \sqrt{\frac{Y A l}{m}}$
C

$f = \frac{1}{2 π} \sqrt{\frac{Y A}{m l}}$
D

$f = \frac{1}{2 π} \sqrt{\frac{m}{Y A l}}$

Solution

For a body under stress we know that its Young’s modulus can be written as $Y = \frac{F l}{A Δ l}$ . Here $\frac{F}{A}$ is stress and $\frac{Δ l}{l}$ is strain. This restoring force can be also written as $F = - k Δ l$ . Hence, we get $k = \frac{F}{Δ l} = \frac{Y A}{l}$ . We know that for a simple harmonic motion the frequency of oscillation is given by $f = \frac{1}{2 π} \sqrt{\frac{k}{m}}$ . Substituting for the value of k we get $f = \frac{1}{2 π} \sqrt{\frac{Y A}{m l}}$ .

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