JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Elasticity & SHM

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A body is performing simple harmonic with an amplitude of 10 cm . The velocity of the body was tripled by air Jet when it is at 5 cm from its mean position. The new amplitude of vibration is

x

cm. The value of x is ______.

Answer: 700

Solution

Initial amplitude

A = 10

cm. Total energy:

E = \frac{1}{2} k A^{2} = \frac{1}{2} k (100)

.

At

x = 5

cm, using energy conservation:

\frac{1}{2} k (100) = \frac{1}{2} k (25) + \frac{1}{2} m v^{2}

v = \frac{75 k}{m}

When velocity is tripled to

3 v = 3 \frac{75 k}{m}

, the potential energy at

x = 5

cm remains the same, but kinetic energy increases:

E_{n e w} = \frac{1}{2} k (25) + \frac{1}{2} m (3 v)^{2} = \frac{1}{2} k (25) + 9 \times \frac{1}{2} m (\frac{75 k}{m})

E_{n e w} = \frac{1}{2} k (25) + \frac{1}{2} k (675) = \frac{1}{2} k (700)

New amplitude:

A_{n e w} = 700

cm

Therefore,

x = 700

JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Elasticity & SHM

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