8. The distance between adjacent nodes in a standing wave pattern in a length of string is 25.0 cm a. What is the wavelength of wave in the string? b. If the frequency of vibration is 200 Hz, calculate the velocity of the wave. 9. Standing waves are produced in a string by sources at each end with a frequency of 10.0 Hz The distance between the third and sixth node is 54 cm. a. What is the wavelength of the interfering waves? b. What is the speed of the waves?

Let's solve each question step by step.<br /><br />### Question 8<br /><br />**Given:**<br />- Distance between adjacent nodes = \(25.0 \, \text{cm}\)<br /><br />#### a. What is the wavelength of wave in the string?<br /><br />In a standing wave pattern on a string, the distance between two consecutive nodes (or antinodes) is half of the wavelength (\(\lambda\)). Therefore:<br /><br />\[<br />\frac{\lambda}{2} = 25.0 \, \text{cm}<br />\]<br /><br />To find the full wavelength:<br /><br />\[<br />\lambda = 2 \times 25.0\, cm = 50.0\, cm<br />= 0.5\, m <br />(1m=100cm)<br />\\<br />So,<br />Wavelength (\(\lambda)\) ≈ **50 cm or .5 meters**<br /><br />---<br /><br />#### b. If frequency of vibration is \(200 Hz,\) calculate velocity of wave.<br /><br />The formula for calculating wave speed (\(v)\) using frequency and wavelength is given by:<br />\[ <br />v = f × λ <br />\\<br />Where:<br />f - Frequency (Hz),<br />λ - Wavelength (meters).<br />\\ <br /><br />Substituting values into this equation gives us,<br /><br />\( v=\left(200\:Hz\times .5\:m)=100\:ms^{-1}. \\<br /><br /><br />Wave Velocity(v): ≈ **100 m/s**<br /><br />---<br /><br /><br /><br />### Question9<br /><br />**Given:**<br />- Frequency \(f=10.\!00 Hz;\)<br />- The distance between third node and sixth node equals to \(54.cm; \\)<br /><br /><br />We know that there are three segments from Node3 to Node6.<br />Thus we can say;<br /><br />Distance Between Nodes(n,m):<br />d=(n-m)*distance_between_adjacent_nodes;<br />where n>m.<br /><br /><br />Here d represents total length covered which means it will be equal to number_of_segments*half_wavelenght;<br /><br /><br />Let’s denote as follows :<br />Number Of Segments=n-m=>3 ; <br />Total Length=d =>54.cm ;<br />Half_WaveLength=hwl ;<br /><br />From above relation :<br /><br />hwl * Number_Of_Segments=d <br /><br />Now substituing known value : <br /><br /><br />hwl×3=m→ hwl=(d/numberofsegments); → ⇒ h wl =(54/cm)/3⇒18./c.m.; <br /><br /><br />Since Half Wave Length(hwL )is calculated , now let’s get Full wavelenght(wl):<br /><br />Full Wavlength(WL)=2*hWL; <br /> WL≈36.c.m.(which converts back again into meter form). <br /><br />Therefore final answer would yield:<br /><br /><br />a.Wavelength:\approx36.CM.or..036.Meters.<br /><br /><br /><br />---<br /> <br />b.What Is Speed Of Waves? <br /><br />Using same previous relationship where V=f*wvl;<br /><br />V=speed(frequency)(wave_length);<br /><br />substituting our previously found results yields:<br /><br /><br /><br />Speed(V)=Frequency*f x WVL=wavespeed=v.f.wl=<br /><br />10.Hz*(o.o36)m=.360.ms^-1.<br /><br /><br /><br /><br />Final Answer For Part B Would Yield Approximately As Follows:<br /><br /><br />Speed(V):≅***O.O360 M/S***<br /><br />