- A$25$
- B$50$
- ✓$75$
- D$100$
Mass per unit length of the wire.
$\mu=4 \times 10^{-2} \mathrm{\,kg} \mathrm{m}^{-1}$
$\therefore $ length of the wire, $L=\frac{M}{\mu}$
$=\frac{30 \times 10^{-3} \mathrm{\,kg}}{4 \times 10^{-2} \mathrm{\,kgm}^{-1}}=0.75 \mathrm{\,m}$
For the fundamental mode $\frac{\lambda}{2}=L$
$\Rightarrow \lambda=2 \mathrm{L}=2 \times 0.75=1.5 \mathrm{\,m}$
Speed of the transverse wave.
$\mathrm{v}=\mathrm{n} \lambda=\left(50 \mathrm{\,s}^{-1}\right)(1.5 \mathrm{\,m})=75 \mathrm{\,ms}^{-1}$
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$1.$ If the total energy of the particle is $E$, it will perform periodic motion only if
$(A)$ $E$ $<0$ $(B)$ $E$ $>0$ $(C)$ $\mathrm{V}_0 > \mathrm{E}>0$ $(D)$ $E > V_0$
$2.$ For periodic motion of small amplitude $\mathrm{A}$, the time period $\mathrm{T}$ of this particle is proportional to
$(A)$ $\mathrm{A} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(B)$ $\frac{1}{\mathrm{~A}} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(C)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$ $(D)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$
$3.$ The acceleration of this particle for $|\mathrm{x}|>\mathrm{X}_0$ is
$(A)$ proportional to $\mathrm{V}_0$
$(B)$ proportional to $\frac{\mathrm{V}_0}{\mathrm{mX}_0}$
$(C)$ proportional to $\sqrt{\frac{\mathrm{V}_0}{\mathrm{mX}_0}}$
$(D)$ zero
Give the answer qustion $1,2$ and $3.$