Simple harmonic motion

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Question

A particle of mass 10 g is placed in- la- potential field given by $V = (50 x^{2} + 100)$ J/kg. If the frequency of oscillation is found to be $\frac{x}{π}$ cycle/s, the value of x is _______.

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Solution

Given that potential energy is $U = mV$
$\begin{array}{l} \Rightarrow U = (50 x^{2} + 100) 10^{- 2} \\ F = - \frac{dU}{dx} = - (100 x) 10^{- 2} \\ \Rightarrow {mω}^{2} x = - (100 \times 10^{- 2}) x \\ 10 \times 10^{- 3} ω^{2} x = 100 \times 10^{- 2} x \\ \Rightarrow ω^{2} = 100 \Rightarrow ω = 10 rad / s \\ \Rightarrow f = \frac{ω}{2 π} = \frac{10}{π} = \frac{5}{π} \end{array}$
Hence x = 5.

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Simple harmonic motion

Remember concepts with our Masterclasses.

A particle of mass 10 g is placed in- la- potential field given by V=50x2+100 J/kg. If the frequency of oscillation is found to be xπcycle/s, the value of x is _______.

Given that potential energy is U=mV⇒ U=50x2+10010−2F=−dUdx=−(100x)10−2⇒ mω2x=−100×10−2x10×10−3ω2x=100×10−2x⇒ ω2=100⇒ ω=10rad/s⇒ f=ω2π=10π=5πHence x = 5.

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A particle of mass 10 g is placed in- la- potential field given by $V = (50 x^{2} + 100)$ J/kg. If the frequency of oscillation is found to be $\frac{x}{π}$ cycle/s, the value of x is _______.