# IBDP Physics Unit 4. Waves – Unit 4.5 – Standing waves: IB Style Question Bank HL Paper 2

### Question

One end of a string is attached to an oscillator and the other is fixed to a wall. When the frequency of the oscillator is 360 Hz the standing wave shown is formed on the string. Point X (not shown) is a point on the string at a distance of 10 cm from the oscillator.

(i) State the number of all other points on the string that have the same amplitude and phase as X. 

(ii )The frequency of the oscillator is reduced to 120 Hz. On the diagram, draw the standing wave that will be formed on the string. Ans

i 3 «points»

ii first harmonic mode drawn ### Question

On a guitar, the strings played vibrate between two fixed points. The frequency of vibration is modified by changing the string length using a finger. The different strings have different wave speeds. When a string is plucked, a standing wave forms between the bridge and the finger. (a) Outline how a standing wave is produced on the string.              

(b) The string is displaced 0.4 cm at point P to sound the guitar. Point P on the string vibrates with simple harmonic motion (shm) in its first harmonic with              a frequency of 195 Hz. The sounding length of the string is 62 cm.

(i) Show that the speed of the wave on the string is about 240 m .      

(ii) Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position. (iii) Calculate, in m , the maximum velocity of vibration of point P when it is vibrating with a frequency of 195 Hz.     

(iv) Calculate, in terms of g, the maximum acceleration of P.    

(v) Estimate the displacement needed to double the energy of the string.                  

(c) The string is made to vibrate in its third harmonic. State the distance between consecutive nodes.    

Ans:

a «travelling» wave moves along the length of the string and reflects «at fixed end»  superposition/interference of incident and reflected waves  the superposition of the reflections is reinforced only for certain wavelengths

b i λ = 2 l = 2 × 0.62 = «1.24 m»  v = f λ = 195 ×1.24 = 242 «m s-1»  Answer must be to 3 or more sf or working shown for MP2

b ii straight line through origin with negative gradient

b iii max velocity occurs at x  0 = V= «(2π)(195) $$\sqrt{0.004^{2}}$$ » = 4.9«ms-1 »

b iv a = (2π 195)2 × 0.004 = 6005 «m s−2 »  = 600g  v use of E =A 2 RO Xo2 A = 0.4  = 0.57 «cm» ≅ 0.6 «cm»

c  $$\frac{62}{3}$$ = 21«cm»