A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, $y(x$, $t )=(0.01 \ m ) \sin \left[\left(62.8 \ m ^{-1}\right) x \right] \cos \left[\left(628 s ^{-1}\right) t \right]$. Assuming $\pi=3.14$, the correct statement$(s)$ is (are) :

$(A)$ The number of nodes is $5$ .

$(B)$ The length of the string is $0.25 \ m$.

$(C)$ The maximum displacement of the midpoint of the string its equilibrium position is $0.01 \ m$.

$(D)$ The fundamental frequency is $100 \ Hz$.

A $20 \mathrm{~cm}$ long string, having a mass of $1.0 \mathrm{~g}$, is fixed at both the ends. The tension in the string is $0.5 \mathrm{~N}$. The string is set into vibrations using an external vibrator of frequency $100 \mathrm{~Hz}$. Find the separation (in $cm$) between the successive nodes on the string.

A composition string is made up by joining two strings of different masses per unit length $\rightarrow \mu$ and $4\mu$ . The composite string is under the same tension. A transverse wave pulse $: Y = (6 mm) \,\,sin\,\,(5t + 40x),$ where $‘t’$ is in seconds and $‘x’$ in meters, is sent along the lighter string towards the joint. The joint is at $x = 0$. The equation of the wave pulse reflected from the joint is

A string of length $1 \mathrm{~m}$ and mass $2 \times 10^{-5} \mathrm{~kg}$ is under tension $\mathrm{T}$. when the string vibrates, two successive harmonics are found to occur at frequencies $750 \mathrm{~Hz}$ and $1000 \mathrm{~Hz}$. The value of tension $\mathrm{T}$ is. . . . . . .Newton.

A string of mass $2.50 \;kg$ is under a tension of $200\; N$. The length of the stretched string is $20.0 \;m$. If the transverse jerk is struck at one end of the string, how long (in $sec$) does the disturbance take to reach the other end?

If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is :